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Hawk28
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spider-sql-prompts

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train · 8.66k rows

db_name stringclasses 146 values	prompt stringlengths 310 4.81k
epinions_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # item ( i_id, title ) # review ( a_id, u_id, i_id, rating, rank ) # useracct ( u_id, name ) # trust ( source_u_id, target_u_id, trust ) # # review.i_id can be joined with item.i_id # review.u_id can be joined with useracct.u_id # trust.target_u_id can be joined with useracct.u_id # trust.source_u_id can be joined with useracct.u_id # ### Question: # # Find the titles of items whose rating is higher than the average review rating of all items. # ### SQL: # # SELECT T1.title FROM item AS T1 JOIN review AS T2 ON T1.i_id = T2.i_id WHERE T2.rating > (SELECT avg(rating) FROM review) # ### End.
epinions_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # item ( i_id, title ) # review ( a_id, u_id, i_id, rating, rank ) # useracct ( u_id, name ) # trust ( source_u_id, target_u_id, trust ) # # review.i_id can be joined with item.i_id # review.u_id can be joined with useracct.u_id # trust.target_u_id can be joined with useracct.u_id # trust.source_u_id can be joined with useracct.u_id # ### Question: # # Find the titles of items that received any rating below 5. # ### SQL: # # SELECT T1.title FROM item AS T1 JOIN review AS T2 ON T1.i_id = T2.i_id WHERE T2.rating < 5 # ### End.
epinions_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # item ( i_id, title ) # review ( a_id, u_id, i_id, rating, rank ) # useracct ( u_id, name ) # trust ( source_u_id, target_u_id, trust ) # # review.i_id can be joined with item.i_id # review.u_id can be joined with useracct.u_id # trust.target_u_id can be joined with useracct.u_id # trust.source_u_id can be joined with useracct.u_id # ### Question: # # Find the titles of items that received both a rating higher than 8 and a rating below 5. # ### SQL: # # SELECT T1.title FROM item AS T1 JOIN review AS T2 ON T1.i_id = T2.i_id WHERE T2.rating > 8 INTERSECT SELECT T1.title FROM item AS T1 JOIN review AS T2 ON T1.i_id = T2.i_id WHERE T2.rating < 5 # ### End.
epinions_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # item ( i_id, title ) # review ( a_id, u_id, i_id, rating, rank ) # useracct ( u_id, name ) # trust ( source_u_id, target_u_id, trust ) # # review.i_id can be joined with item.i_id # review.u_id can be joined with useracct.u_id # trust.target_u_id can be joined with useracct.u_id # trust.source_u_id can be joined with useracct.u_id # ### Question: # # Find the names of items whose rank is higher than 3 and whose average rating is above 5. # ### SQL: # # SELECT T1.title FROM item AS T1 JOIN review AS T2 ON T1.i_id = T2.i_id WHERE T2.rank > 3 INTERSECT SELECT T1.title FROM item AS T1 JOIN review AS T2 ON T1.i_id = T2.i_id GROUP BY T2.i_id HAVING avg(T2.rating) > 5 # ### End.
epinions_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # item ( i_id, title ) # review ( a_id, u_id, i_id, rating, rank ) # useracct ( u_id, name ) # trust ( source_u_id, target_u_id, trust ) # # review.i_id can be joined with item.i_id # review.u_id can be joined with useracct.u_id # trust.target_u_id can be joined with useracct.u_id # trust.source_u_id can be joined with useracct.u_id # ### Question: # # Find the name of the item with the lowest average rating. # ### SQL: # # SELECT T1.title FROM item AS T1 JOIN review AS T2 ON T1.i_id = T2.i_id GROUP BY T2.i_id ORDER BY avg(T2.rating) LIMIT 1 # ### End.
epinions_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # item ( i_id, title ) # review ( a_id, u_id, i_id, rating, rank ) # useracct ( u_id, name ) # trust ( source_u_id, target_u_id, trust ) # # review.i_id can be joined with item.i_id # review.u_id can be joined with useracct.u_id # trust.target_u_id can be joined with useracct.u_id # trust.source_u_id can be joined with useracct.u_id # ### Question: # # List the titles of all items in alphabetic order . # ### SQL: # # SELECT title FROM item ORDER BY title # ### End.
epinions_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # item ( i_id, title ) # review ( a_id, u_id, i_id, rating, rank ) # useracct ( u_id, name ) # trust ( source_u_id, target_u_id, trust ) # # review.i_id can be joined with item.i_id # review.u_id can be joined with useracct.u_id # trust.target_u_id can be joined with useracct.u_id # trust.source_u_id can be joined with useracct.u_id # ### Question: # # Find the name of the user who gives the most reviews. # ### SQL: # # SELECT T1.name FROM useracct AS T1 JOIN review AS T2 ON T1.u_id = T2.u_id GROUP BY T2.u_id ORDER BY count(*) DESC LIMIT 1 # ### End.
epinions_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # item ( i_id, title ) # review ( a_id, u_id, i_id, rating, rank ) # useracct ( u_id, name ) # trust ( source_u_id, target_u_id, trust ) # # review.i_id can be joined with item.i_id # review.u_id can be joined with useracct.u_id # trust.target_u_id can be joined with useracct.u_id # trust.source_u_id can be joined with useracct.u_id # ### Question: # # Find the name and id of the item with the highest average rating. # ### SQL: # # SELECT T1.title , T1.i_id FROM item AS T1 JOIN review AS T2 ON T1.i_id = T2.i_id GROUP BY T2.i_id ORDER BY avg(T2.rating) DESC LIMIT 1 # ### End.
epinions_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # item ( i_id, title ) # review ( a_id, u_id, i_id, rating, rank ) # useracct ( u_id, name ) # trust ( source_u_id, target_u_id, trust ) # # review.i_id can be joined with item.i_id # review.u_id can be joined with useracct.u_id # trust.target_u_id can be joined with useracct.u_id # trust.source_u_id can be joined with useracct.u_id # ### Question: # # Find the name and id of the good with the highest average rank. # ### SQL: # # SELECT T1.title , T1.i_id FROM item AS T1 JOIN review AS T2 ON T1.i_id = T2.i_id GROUP BY T2.i_id ORDER BY avg(T2.rank) DESC LIMIT 1 # ### End.
epinions_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # item ( i_id, title ) # review ( a_id, u_id, i_id, rating, rank ) # useracct ( u_id, name ) # trust ( source_u_id, target_u_id, trust ) # # review.i_id can be joined with item.i_id # review.u_id can be joined with useracct.u_id # trust.target_u_id can be joined with useracct.u_id # trust.source_u_id can be joined with useracct.u_id # ### Question: # # For each user, return the name and the average rating of reviews given by them. # ### SQL: # # SELECT T1.name , avg(T2.rating) FROM useracct AS T1 JOIN review AS T2 ON T1.u_id = T2.u_id GROUP BY T2.u_id # ### End.
epinions_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # item ( i_id, title ) # review ( a_id, u_id, i_id, rating, rank ) # useracct ( u_id, name ) # trust ( source_u_id, target_u_id, trust ) # # review.i_id can be joined with item.i_id # review.u_id can be joined with useracct.u_id # trust.target_u_id can be joined with useracct.u_id # trust.source_u_id can be joined with useracct.u_id # ### Question: # # For each user, find their name and the number of reviews written by them. # ### SQL: # # SELECT T1.name , count(*) FROM useracct AS T1 JOIN review AS T2 ON T1.u_id = T2.u_id GROUP BY T2.u_id # ### End.
epinions_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # item ( i_id, title ) # review ( a_id, u_id, i_id, rating, rank ) # useracct ( u_id, name ) # trust ( source_u_id, target_u_id, trust ) # # review.i_id can be joined with item.i_id # review.u_id can be joined with useracct.u_id # trust.target_u_id can be joined with useracct.u_id # trust.source_u_id can be joined with useracct.u_id # ### Question: # # Find the name of the user who gave the highest rating. # ### SQL: # # SELECT T1.name FROM useracct AS T1 JOIN review AS T2 ON T1.u_id = T2.u_id ORDER BY T2.rating DESC LIMIT 1 # ### End.
epinions_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # item ( i_id, title ) # review ( a_id, u_id, i_id, rating, rank ) # useracct ( u_id, name ) # trust ( source_u_id, target_u_id, trust ) # # review.i_id can be joined with item.i_id # review.u_id can be joined with useracct.u_id # trust.target_u_id can be joined with useracct.u_id # trust.source_u_id can be joined with useracct.u_id # ### Question: # # Find the name of the source user with the highest average trust score. # ### SQL: # # SELECT T1.name FROM useracct AS T1 JOIN trust AS T2 ON T1.u_id = T2.source_u_id GROUP BY T2.source_u_id ORDER BY avg(trust) DESC LIMIT 1 # ### End.
epinions_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # item ( i_id, title ) # review ( a_id, u_id, i_id, rating, rank ) # useracct ( u_id, name ) # trust ( source_u_id, target_u_id, trust ) # # review.i_id can be joined with item.i_id # review.u_id can be joined with useracct.u_id # trust.target_u_id can be joined with useracct.u_id # trust.source_u_id can be joined with useracct.u_id # ### Question: # # Find each target user's name and average trust score. # ### SQL: # # SELECT T1.name , avg(trust) FROM useracct AS T1 JOIN trust AS T2 ON T1.u_id = T2.target_u_id GROUP BY T2.target_u_id # ### End.
epinions_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # item ( i_id, title ) # review ( a_id, u_id, i_id, rating, rank ) # useracct ( u_id, name ) # trust ( source_u_id, target_u_id, trust ) # # review.i_id can be joined with item.i_id # review.u_id can be joined with useracct.u_id # trust.target_u_id can be joined with useracct.u_id # trust.source_u_id can be joined with useracct.u_id # ### Question: # # Find the name of the target user with the lowest trust score. # ### SQL: # # SELECT T1.name FROM useracct AS T1 JOIN trust AS T2 ON T1.u_id = T2.target_u_id ORDER BY trust LIMIT 1 # ### End.
epinions_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # item ( i_id, title ) # review ( a_id, u_id, i_id, rating, rank ) # useracct ( u_id, name ) # trust ( source_u_id, target_u_id, trust ) # # review.i_id can be joined with item.i_id # review.u_id can be joined with useracct.u_id # trust.target_u_id can be joined with useracct.u_id # trust.source_u_id can be joined with useracct.u_id # ### Question: # # Find the names of the items that did not receive any review. # ### SQL: # # SELECT title FROM item WHERE i_id NOT IN (SELECT i_id FROM review) # ### End.
epinions_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # item ( i_id, title ) # review ( a_id, u_id, i_id, rating, rank ) # useracct ( u_id, name ) # trust ( source_u_id, target_u_id, trust ) # # review.i_id can be joined with item.i_id # review.u_id can be joined with useracct.u_id # trust.target_u_id can be joined with useracct.u_id # trust.source_u_id can be joined with useracct.u_id # ### Question: # # Find the names of users who did not leave any review. # ### SQL: # # SELECT name FROM useracct WHERE u_id NOT IN (SELECT u_id FROM review) # ### End.
epinions_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # item ( i_id, title ) # review ( a_id, u_id, i_id, rating, rank ) # useracct ( u_id, name ) # trust ( source_u_id, target_u_id, trust ) # # review.i_id can be joined with item.i_id # review.u_id can be joined with useracct.u_id # trust.target_u_id can be joined with useracct.u_id # trust.source_u_id can be joined with useracct.u_id # ### Question: # # Find the number of users who did not write any review. # ### SQL: # # SELECT count(*) FROM useracct WHERE u_id NOT IN (SELECT u_id FROM review) # ### End.
epinions_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # item ( i_id, title ) # review ( a_id, u_id, i_id, rating, rank ) # useracct ( u_id, name ) # trust ( source_u_id, target_u_id, trust ) # # review.i_id can be joined with item.i_id # review.u_id can be joined with useracct.u_id # trust.target_u_id can be joined with useracct.u_id # trust.source_u_id can be joined with useracct.u_id # ### Question: # # Find the number of items without any review. # ### SQL: # # SELECT count(*) FROM item WHERE i_id NOT IN (SELECT i_id FROM review) # ### End.
riding_club	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # player ( Player_ID, Sponsor_name, Player_name, Gender, Residence, Occupation, Votes, Rank ) # club ( Club_ID, Club_name, Region, Start_year ) # coach ( Coach_ID, Coach_name, Gender, Club_ID, Rank ) # player_coach ( Player_ID, Coach_ID, Starting_year ) # match_result ( Rank, Club_ID, Gold, Big_Silver, Small_Silver, Bronze, Points ) # # coach.Club_ID can be joined with club.Club_ID # player_coach.Coach_ID can be joined with coach.Coach_ID # player_coach.Player_ID can be joined with player.Player_ID # match_result.Club_ID can be joined with club.Club_ID # ### Question: # # How many players are there? # ### SQL: # # SELECT count(*) FROM player # ### End.
riding_club	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # player ( Player_ID, Sponsor_name, Player_name, Gender, Residence, Occupation, Votes, Rank ) # club ( Club_ID, Club_name, Region, Start_year ) # coach ( Coach_ID, Coach_name, Gender, Club_ID, Rank ) # player_coach ( Player_ID, Coach_ID, Starting_year ) # match_result ( Rank, Club_ID, Gold, Big_Silver, Small_Silver, Bronze, Points ) # # coach.Club_ID can be joined with club.Club_ID # player_coach.Coach_ID can be joined with coach.Coach_ID # player_coach.Player_ID can be joined with player.Player_ID # match_result.Club_ID can be joined with club.Club_ID # ### Question: # # List the names of players in ascending order of votes. # ### SQL: # # SELECT Player_name FROM player ORDER BY Votes ASC # ### End.
riding_club	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # player ( Player_ID, Sponsor_name, Player_name, Gender, Residence, Occupation, Votes, Rank ) # club ( Club_ID, Club_name, Region, Start_year ) # coach ( Coach_ID, Coach_name, Gender, Club_ID, Rank ) # player_coach ( Player_ID, Coach_ID, Starting_year ) # match_result ( Rank, Club_ID, Gold, Big_Silver, Small_Silver, Bronze, Points ) # # coach.Club_ID can be joined with club.Club_ID # player_coach.Coach_ID can be joined with coach.Coach_ID # player_coach.Player_ID can be joined with player.Player_ID # match_result.Club_ID can be joined with club.Club_ID # ### Question: # # What are the gender and occupation of players? # ### SQL: # # SELECT Gender , Occupation FROM player # ### End.
riding_club	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # player ( Player_ID, Sponsor_name, Player_name, Gender, Residence, Occupation, Votes, Rank ) # club ( Club_ID, Club_name, Region, Start_year ) # coach ( Coach_ID, Coach_name, Gender, Club_ID, Rank ) # player_coach ( Player_ID, Coach_ID, Starting_year ) # match_result ( Rank, Club_ID, Gold, Big_Silver, Small_Silver, Bronze, Points ) # # coach.Club_ID can be joined with club.Club_ID # player_coach.Coach_ID can be joined with coach.Coach_ID # player_coach.Player_ID can be joined with player.Player_ID # match_result.Club_ID can be joined with club.Club_ID # ### Question: # # List the name and residence for players whose occupation is not "Researcher". # ### SQL: # # SELECT Player_name , residence FROM player WHERE Occupation != "Researcher" # ### End.
riding_club	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # player ( Player_ID, Sponsor_name, Player_name, Gender, Residence, Occupation, Votes, Rank ) # club ( Club_ID, Club_name, Region, Start_year ) # coach ( Coach_ID, Coach_name, Gender, Club_ID, Rank ) # player_coach ( Player_ID, Coach_ID, Starting_year ) # match_result ( Rank, Club_ID, Gold, Big_Silver, Small_Silver, Bronze, Points ) # # coach.Club_ID can be joined with club.Club_ID # player_coach.Coach_ID can be joined with coach.Coach_ID # player_coach.Player_ID can be joined with player.Player_ID # match_result.Club_ID can be joined with club.Club_ID # ### Question: # # Show the names of sponsors of players whose residence is either "Brandon" or "Birtle". # ### SQL: # # SELECT Sponsor_name FROM player WHERE Residence = "Brandon" OR Residence = "Birtle" # ### End.
riding_club	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # player ( Player_ID, Sponsor_name, Player_name, Gender, Residence, Occupation, Votes, Rank ) # club ( Club_ID, Club_name, Region, Start_year ) # coach ( Coach_ID, Coach_name, Gender, Club_ID, Rank ) # player_coach ( Player_ID, Coach_ID, Starting_year ) # match_result ( Rank, Club_ID, Gold, Big_Silver, Small_Silver, Bronze, Points ) # # coach.Club_ID can be joined with club.Club_ID # player_coach.Coach_ID can be joined with coach.Coach_ID # player_coach.Player_ID can be joined with player.Player_ID # match_result.Club_ID can be joined with club.Club_ID # ### Question: # # What is the name of the player with the largest number of votes? # ### SQL: # # SELECT Player_name FROM player ORDER BY Votes DESC LIMIT 1 # ### End.
riding_club	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # player ( Player_ID, Sponsor_name, Player_name, Gender, Residence, Occupation, Votes, Rank ) # club ( Club_ID, Club_name, Region, Start_year ) # coach ( Coach_ID, Coach_name, Gender, Club_ID, Rank ) # player_coach ( Player_ID, Coach_ID, Starting_year ) # match_result ( Rank, Club_ID, Gold, Big_Silver, Small_Silver, Bronze, Points ) # # coach.Club_ID can be joined with club.Club_ID # player_coach.Coach_ID can be joined with coach.Coach_ID # player_coach.Player_ID can be joined with player.Player_ID # match_result.Club_ID can be joined with club.Club_ID # ### Question: # # Show different occupations along with the number of players in each occupation. # ### SQL: # # SELECT Occupation , COUNT(*) FROM player GROUP BY Occupation # ### End.
riding_club	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # player ( Player_ID, Sponsor_name, Player_name, Gender, Residence, Occupation, Votes, Rank ) # club ( Club_ID, Club_name, Region, Start_year ) # coach ( Coach_ID, Coach_name, Gender, Club_ID, Rank ) # player_coach ( Player_ID, Coach_ID, Starting_year ) # match_result ( Rank, Club_ID, Gold, Big_Silver, Small_Silver, Bronze, Points ) # # coach.Club_ID can be joined with club.Club_ID # player_coach.Coach_ID can be joined with coach.Coach_ID # player_coach.Player_ID can be joined with player.Player_ID # match_result.Club_ID can be joined with club.Club_ID # ### Question: # # Please show the most common occupation of players. # ### SQL: # # SELECT Occupation FROM player GROUP BY Occupation ORDER BY COUNT(*) DESC LIMIT 1 # ### End.
riding_club	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # player ( Player_ID, Sponsor_name, Player_name, Gender, Residence, Occupation, Votes, Rank ) # club ( Club_ID, Club_name, Region, Start_year ) # coach ( Coach_ID, Coach_name, Gender, Club_ID, Rank ) # player_coach ( Player_ID, Coach_ID, Starting_year ) # match_result ( Rank, Club_ID, Gold, Big_Silver, Small_Silver, Bronze, Points ) # # coach.Club_ID can be joined with club.Club_ID # player_coach.Coach_ID can be joined with coach.Coach_ID # player_coach.Player_ID can be joined with player.Player_ID # match_result.Club_ID can be joined with club.Club_ID # ### Question: # # Show the residences that have at least two players. # ### SQL: # # SELECT Residence FROM player GROUP BY Residence HAVING COUNT(*) >= 2 # ### End.
riding_club	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # player ( Player_ID, Sponsor_name, Player_name, Gender, Residence, Occupation, Votes, Rank ) # club ( Club_ID, Club_name, Region, Start_year ) # coach ( Coach_ID, Coach_name, Gender, Club_ID, Rank ) # player_coach ( Player_ID, Coach_ID, Starting_year ) # match_result ( Rank, Club_ID, Gold, Big_Silver, Small_Silver, Bronze, Points ) # # coach.Club_ID can be joined with club.Club_ID # player_coach.Coach_ID can be joined with coach.Coach_ID # player_coach.Player_ID can be joined with player.Player_ID # match_result.Club_ID can be joined with club.Club_ID # ### Question: # # Show the names of players and names of their coaches. # ### SQL: # # SELECT T3.Player_name , T2.coach_name FROM player_coach AS T1 JOIN coach AS T2 ON T1.Coach_ID = T2.Coach_ID JOIN player AS T3 ON T1.Player_ID = T3.Player_ID # ### End.
riding_club	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # player ( Player_ID, Sponsor_name, Player_name, Gender, Residence, Occupation, Votes, Rank ) # club ( Club_ID, Club_name, Region, Start_year ) # coach ( Coach_ID, Coach_name, Gender, Club_ID, Rank ) # player_coach ( Player_ID, Coach_ID, Starting_year ) # match_result ( Rank, Club_ID, Gold, Big_Silver, Small_Silver, Bronze, Points ) # # coach.Club_ID can be joined with club.Club_ID # player_coach.Coach_ID can be joined with coach.Coach_ID # player_coach.Player_ID can be joined with player.Player_ID # match_result.Club_ID can be joined with club.Club_ID # ### Question: # # Show the names of players coached by the rank 1 coach. # ### SQL: # # SELECT T3.Player_name FROM player_coach AS T1 JOIN coach AS T2 ON T1.Coach_ID = T2.Coach_ID JOIN player AS T3 ON T1.Player_ID = T3.Player_ID WHERE T2.Rank = 1 # ### End.
riding_club	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # player ( Player_ID, Sponsor_name, Player_name, Gender, Residence, Occupation, Votes, Rank ) # club ( Club_ID, Club_name, Region, Start_year ) # coach ( Coach_ID, Coach_name, Gender, Club_ID, Rank ) # player_coach ( Player_ID, Coach_ID, Starting_year ) # match_result ( Rank, Club_ID, Gold, Big_Silver, Small_Silver, Bronze, Points ) # # coach.Club_ID can be joined with club.Club_ID # player_coach.Coach_ID can be joined with coach.Coach_ID # player_coach.Player_ID can be joined with player.Player_ID # match_result.Club_ID can be joined with club.Club_ID # ### Question: # # Show the names and genders of players with a coach starting after 2011. # ### SQL: # # SELECT T3.Player_name , T3.gender FROM player_coach AS T1 JOIN coach AS T2 ON T1.Coach_ID = T2.Coach_ID JOIN player AS T3 ON T1.Player_ID = T3.Player_ID WHERE T1.Starting_year > 2011 # ### End.
riding_club	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # player ( Player_ID, Sponsor_name, Player_name, Gender, Residence, Occupation, Votes, Rank ) # club ( Club_ID, Club_name, Region, Start_year ) # coach ( Coach_ID, Coach_name, Gender, Club_ID, Rank ) # player_coach ( Player_ID, Coach_ID, Starting_year ) # match_result ( Rank, Club_ID, Gold, Big_Silver, Small_Silver, Bronze, Points ) # # coach.Club_ID can be joined with club.Club_ID # player_coach.Coach_ID can be joined with coach.Coach_ID # player_coach.Player_ID can be joined with player.Player_ID # match_result.Club_ID can be joined with club.Club_ID # ### Question: # # Show the names of players and names of their coaches in descending order of the votes of players. # ### SQL: # # SELECT T3.Player_name , T2.coach_name FROM player_coach AS T1 JOIN coach AS T2 ON T1.Coach_ID = T2.Coach_ID JOIN player AS T3 ON T1.Player_ID = T3.Player_ID ORDER BY T3.Votes DESC # ### End.
riding_club	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # player ( Player_ID, Sponsor_name, Player_name, Gender, Residence, Occupation, Votes, Rank ) # club ( Club_ID, Club_name, Region, Start_year ) # coach ( Coach_ID, Coach_name, Gender, Club_ID, Rank ) # player_coach ( Player_ID, Coach_ID, Starting_year ) # match_result ( Rank, Club_ID, Gold, Big_Silver, Small_Silver, Bronze, Points ) # # coach.Club_ID can be joined with club.Club_ID # player_coach.Coach_ID can be joined with coach.Coach_ID # player_coach.Player_ID can be joined with player.Player_ID # match_result.Club_ID can be joined with club.Club_ID # ### Question: # # List the names of players that do not have coaches. # ### SQL: # # SELECT Player_name FROM player WHERE Player_ID NOT IN (SELECT Player_ID FROM player_coach) # ### End.
riding_club	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # player ( Player_ID, Sponsor_name, Player_name, Gender, Residence, Occupation, Votes, Rank ) # club ( Club_ID, Club_name, Region, Start_year ) # coach ( Coach_ID, Coach_name, Gender, Club_ID, Rank ) # player_coach ( Player_ID, Coach_ID, Starting_year ) # match_result ( Rank, Club_ID, Gold, Big_Silver, Small_Silver, Bronze, Points ) # # coach.Club_ID can be joined with club.Club_ID # player_coach.Coach_ID can be joined with coach.Coach_ID # player_coach.Player_ID can be joined with player.Player_ID # match_result.Club_ID can be joined with club.Club_ID # ### Question: # # Show the residences that have both a player of gender "M" and a player of gender "F". # ### SQL: # # SELECT Residence FROM player WHERE gender = "M" INTERSECT SELECT Residence FROM player WHERE gender = "F" # ### End.
riding_club	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # player ( Player_ID, Sponsor_name, Player_name, Gender, Residence, Occupation, Votes, Rank ) # club ( Club_ID, Club_name, Region, Start_year ) # coach ( Coach_ID, Coach_name, Gender, Club_ID, Rank ) # player_coach ( Player_ID, Coach_ID, Starting_year ) # match_result ( Rank, Club_ID, Gold, Big_Silver, Small_Silver, Bronze, Points ) # # coach.Club_ID can be joined with club.Club_ID # player_coach.Coach_ID can be joined with coach.Coach_ID # player_coach.Player_ID can be joined with player.Player_ID # match_result.Club_ID can be joined with club.Club_ID # ### Question: # # How many coaches does each club has? List the club id, name and the number of coaches. # ### SQL: # # SELECT T1.club_id , T1.club_name, count(*) FROM club AS T1 JOIN coach AS T2 ON T1.club_id = T2.club_id GROUP BY T1.club_id # ### End.
riding_club	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # player ( Player_ID, Sponsor_name, Player_name, Gender, Residence, Occupation, Votes, Rank ) # club ( Club_ID, Club_name, Region, Start_year ) # coach ( Coach_ID, Coach_name, Gender, Club_ID, Rank ) # player_coach ( Player_ID, Coach_ID, Starting_year ) # match_result ( Rank, Club_ID, Gold, Big_Silver, Small_Silver, Bronze, Points ) # # coach.Club_ID can be joined with club.Club_ID # player_coach.Coach_ID can be joined with coach.Coach_ID # player_coach.Player_ID can be joined with player.Player_ID # match_result.Club_ID can be joined with club.Club_ID # ### Question: # # How many gold medals has the club with the most coaches won? # ### SQL: # # SELECT T1.club_id , T1.gold FROM match_result AS T1 JOIN coach AS T2 ON T1.club_id = T2.club_id GROUP BY T1.club_id ORDER BY count(*) DESC LIMIT 1 # ### End.
gymnast	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # gymnast ( Gymnast_ID, Floor_Exercise_Points, Pommel_Horse_Points, Rings_Points, Vault_Points, Parallel_Bars_Points, Horizontal_Bar_Points, Total_Points ) # people ( People_ID, Name, Age, Height, Hometown ) # # gymnast.Gymnast_ID can be joined with people.People_ID # ### Question: # # How many gymnasts are there? # ### SQL: # # SELECT count(*) FROM gymnast # ### End.
gymnast	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # gymnast ( Gymnast_ID, Floor_Exercise_Points, Pommel_Horse_Points, Rings_Points, Vault_Points, Parallel_Bars_Points, Horizontal_Bar_Points, Total_Points ) # people ( People_ID, Name, Age, Height, Hometown ) # # gymnast.Gymnast_ID can be joined with people.People_ID # ### Question: # # Count the number of gymnasts. # ### SQL: # # SELECT count(*) FROM gymnast # ### End.
gymnast	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # gymnast ( Gymnast_ID, Floor_Exercise_Points, Pommel_Horse_Points, Rings_Points, Vault_Points, Parallel_Bars_Points, Horizontal_Bar_Points, Total_Points ) # people ( People_ID, Name, Age, Height, Hometown ) # # gymnast.Gymnast_ID can be joined with people.People_ID # ### Question: # # List the total points of gymnasts in descending order. # ### SQL: # # SELECT Total_Points FROM gymnast ORDER BY Total_Points DESC # ### End.
gymnast	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # gymnast ( Gymnast_ID, Floor_Exercise_Points, Pommel_Horse_Points, Rings_Points, Vault_Points, Parallel_Bars_Points, Horizontal_Bar_Points, Total_Points ) # people ( People_ID, Name, Age, Height, Hometown ) # # gymnast.Gymnast_ID can be joined with people.People_ID # ### Question: # # What are the total points for all gymnasts, ordered by total points descending? # ### SQL: # # SELECT Total_Points FROM gymnast ORDER BY Total_Points DESC # ### End.
gymnast	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # gymnast ( Gymnast_ID, Floor_Exercise_Points, Pommel_Horse_Points, Rings_Points, Vault_Points, Parallel_Bars_Points, Horizontal_Bar_Points, Total_Points ) # people ( People_ID, Name, Age, Height, Hometown ) # # gymnast.Gymnast_ID can be joined with people.People_ID # ### Question: # # List the total points of gymnasts in descending order of floor exercise points. # ### SQL: # # SELECT Total_Points FROM gymnast ORDER BY Floor_Exercise_Points DESC # ### End.
gymnast	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # gymnast ( Gymnast_ID, Floor_Exercise_Points, Pommel_Horse_Points, Rings_Points, Vault_Points, Parallel_Bars_Points, Horizontal_Bar_Points, Total_Points ) # people ( People_ID, Name, Age, Height, Hometown ) # # gymnast.Gymnast_ID can be joined with people.People_ID # ### Question: # # What are the total points of gymnasts, ordered by their floor exercise points descending? # ### SQL: # # SELECT Total_Points FROM gymnast ORDER BY Floor_Exercise_Points DESC # ### End.
gymnast	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # gymnast ( Gymnast_ID, Floor_Exercise_Points, Pommel_Horse_Points, Rings_Points, Vault_Points, Parallel_Bars_Points, Horizontal_Bar_Points, Total_Points ) # people ( People_ID, Name, Age, Height, Hometown ) # # gymnast.Gymnast_ID can be joined with people.People_ID # ### Question: # # What is the average horizontal bar points for all gymnasts? # ### SQL: # # SELECT avg(Horizontal_Bar_Points) FROM gymnast # ### End.
gymnast	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # gymnast ( Gymnast_ID, Floor_Exercise_Points, Pommel_Horse_Points, Rings_Points, Vault_Points, Parallel_Bars_Points, Horizontal_Bar_Points, Total_Points ) # people ( People_ID, Name, Age, Height, Hometown ) # # gymnast.Gymnast_ID can be joined with people.People_ID # ### Question: # # Return the average horizontal bar points across all gymnasts. # ### SQL: # # SELECT avg(Horizontal_Bar_Points) FROM gymnast # ### End.
gymnast	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # gymnast ( Gymnast_ID, Floor_Exercise_Points, Pommel_Horse_Points, Rings_Points, Vault_Points, Parallel_Bars_Points, Horizontal_Bar_Points, Total_Points ) # people ( People_ID, Name, Age, Height, Hometown ) # # gymnast.Gymnast_ID can be joined with people.People_ID # ### Question: # # What are the names of people in ascending alphabetical order? # ### SQL: # # SELECT Name FROM People ORDER BY Name ASC # ### End.
gymnast	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # gymnast ( Gymnast_ID, Floor_Exercise_Points, Pommel_Horse_Points, Rings_Points, Vault_Points, Parallel_Bars_Points, Horizontal_Bar_Points, Total_Points ) # people ( People_ID, Name, Age, Height, Hometown ) # # gymnast.Gymnast_ID can be joined with people.People_ID # ### Question: # # Return the names of people, ordered alphabetically. # ### SQL: # # SELECT Name FROM People ORDER BY Name ASC # ### End.
gymnast	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # gymnast ( Gymnast_ID, Floor_Exercise_Points, Pommel_Horse_Points, Rings_Points, Vault_Points, Parallel_Bars_Points, Horizontal_Bar_Points, Total_Points ) # people ( People_ID, Name, Age, Height, Hometown ) # # gymnast.Gymnast_ID can be joined with people.People_ID # ### Question: # # What are the names of gymnasts? # ### SQL: # # SELECT T2.Name FROM gymnast AS T1 JOIN people AS T2 ON T1.Gymnast_ID = T2.People_ID # ### End.
gymnast	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # gymnast ( Gymnast_ID, Floor_Exercise_Points, Pommel_Horse_Points, Rings_Points, Vault_Points, Parallel_Bars_Points, Horizontal_Bar_Points, Total_Points ) # people ( People_ID, Name, Age, Height, Hometown ) # # gymnast.Gymnast_ID can be joined with people.People_ID # ### Question: # # Return the names of the gymnasts. # ### SQL: # # SELECT T2.Name FROM gymnast AS T1 JOIN people AS T2 ON T1.Gymnast_ID = T2.People_ID # ### End.
gymnast	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # gymnast ( Gymnast_ID, Floor_Exercise_Points, Pommel_Horse_Points, Rings_Points, Vault_Points, Parallel_Bars_Points, Horizontal_Bar_Points, Total_Points ) # people ( People_ID, Name, Age, Height, Hometown ) # # gymnast.Gymnast_ID can be joined with people.People_ID # ### Question: # # What are the names of gymnasts whose hometown is not "Santo Domingo"? # ### SQL: # # SELECT T2.Name FROM gymnast AS T1 JOIN people AS T2 ON T1.Gymnast_ID = T2.People_ID WHERE T2.Hometown != "Santo Domingo" # ### End.
gymnast	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # gymnast ( Gymnast_ID, Floor_Exercise_Points, Pommel_Horse_Points, Rings_Points, Vault_Points, Parallel_Bars_Points, Horizontal_Bar_Points, Total_Points ) # people ( People_ID, Name, Age, Height, Hometown ) # # gymnast.Gymnast_ID can be joined with people.People_ID # ### Question: # # Return the names of gymnasts who did not grow up in Santo Domingo. # ### SQL: # # SELECT T2.Name FROM gymnast AS T1 JOIN people AS T2 ON T1.Gymnast_ID = T2.People_ID WHERE T2.Hometown != "Santo Domingo" # ### End.
gymnast	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # gymnast ( Gymnast_ID, Floor_Exercise_Points, Pommel_Horse_Points, Rings_Points, Vault_Points, Parallel_Bars_Points, Horizontal_Bar_Points, Total_Points ) # people ( People_ID, Name, Age, Height, Hometown ) # # gymnast.Gymnast_ID can be joined with people.People_ID # ### Question: # # What is the age of the tallest person? # ### SQL: # # SELECT Age FROM people ORDER BY Height DESC LIMIT 1 # ### End.
gymnast	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # gymnast ( Gymnast_ID, Floor_Exercise_Points, Pommel_Horse_Points, Rings_Points, Vault_Points, Parallel_Bars_Points, Horizontal_Bar_Points, Total_Points ) # people ( People_ID, Name, Age, Height, Hometown ) # # gymnast.Gymnast_ID can be joined with people.People_ID # ### Question: # # Return the age of the person with the greatest height. # ### SQL: # # SELECT Age FROM people ORDER BY Height DESC LIMIT 1 # ### End.
gymnast	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # gymnast ( Gymnast_ID, Floor_Exercise_Points, Pommel_Horse_Points, Rings_Points, Vault_Points, Parallel_Bars_Points, Horizontal_Bar_Points, Total_Points ) # people ( People_ID, Name, Age, Height, Hometown ) # # gymnast.Gymnast_ID can be joined with people.People_ID # ### Question: # # List the names of the top 5 oldest people. # ### SQL: # # SELECT Name FROM People ORDER BY Age DESC LIMIT 5 # ### End.
gymnast	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # gymnast ( Gymnast_ID, Floor_Exercise_Points, Pommel_Horse_Points, Rings_Points, Vault_Points, Parallel_Bars_Points, Horizontal_Bar_Points, Total_Points ) # people ( People_ID, Name, Age, Height, Hometown ) # # gymnast.Gymnast_ID can be joined with people.People_ID # ### Question: # # What are the names of the five oldest people? # ### SQL: # # SELECT Name FROM People ORDER BY Age DESC LIMIT 5 # ### End.
gymnast	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # gymnast ( Gymnast_ID, Floor_Exercise_Points, Pommel_Horse_Points, Rings_Points, Vault_Points, Parallel_Bars_Points, Horizontal_Bar_Points, Total_Points ) # people ( People_ID, Name, Age, Height, Hometown ) # # gymnast.Gymnast_ID can be joined with people.People_ID # ### Question: # # What is the total point count of the youngest gymnast? # ### SQL: # # SELECT T1.Total_Points FROM gymnast AS T1 JOIN people AS T2 ON T1.Gymnast_ID = T2.People_ID ORDER BY T2.Age ASC LIMIT 1 # ### End.
gymnast	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # gymnast ( Gymnast_ID, Floor_Exercise_Points, Pommel_Horse_Points, Rings_Points, Vault_Points, Parallel_Bars_Points, Horizontal_Bar_Points, Total_Points ) # people ( People_ID, Name, Age, Height, Hometown ) # # gymnast.Gymnast_ID can be joined with people.People_ID # ### Question: # # Return the total points of the gymnast with the lowest age. # ### SQL: # # SELECT T1.Total_Points FROM gymnast AS T1 JOIN people AS T2 ON T1.Gymnast_ID = T2.People_ID ORDER BY T2.Age ASC LIMIT 1 # ### End.
gymnast	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # gymnast ( Gymnast_ID, Floor_Exercise_Points, Pommel_Horse_Points, Rings_Points, Vault_Points, Parallel_Bars_Points, Horizontal_Bar_Points, Total_Points ) # people ( People_ID, Name, Age, Height, Hometown ) # # gymnast.Gymnast_ID can be joined with people.People_ID # ### Question: # # What is the average age of all gymnasts? # ### SQL: # # SELECT avg(T2.Age) FROM gymnast AS T1 JOIN people AS T2 ON T1.Gymnast_ID = T2.People_ID # ### End.
gymnast	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # gymnast ( Gymnast_ID, Floor_Exercise_Points, Pommel_Horse_Points, Rings_Points, Vault_Points, Parallel_Bars_Points, Horizontal_Bar_Points, Total_Points ) # people ( People_ID, Name, Age, Height, Hometown ) # # gymnast.Gymnast_ID can be joined with people.People_ID # ### Question: # # Return the average age across all gymnasts. # ### SQL: # # SELECT avg(T2.Age) FROM gymnast AS T1 JOIN people AS T2 ON T1.Gymnast_ID = T2.People_ID # ### End.
gymnast	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # gymnast ( Gymnast_ID, Floor_Exercise_Points, Pommel_Horse_Points, Rings_Points, Vault_Points, Parallel_Bars_Points, Horizontal_Bar_Points, Total_Points ) # people ( People_ID, Name, Age, Height, Hometown ) # # gymnast.Gymnast_ID can be joined with people.People_ID # ### Question: # # What are the distinct hometowns of gymnasts with total points more than 57.5? # ### SQL: # # SELECT DISTINCT T2.Hometown FROM gymnast AS T1 JOIN people AS T2 ON T1.Gymnast_ID = T2.People_ID WHERE T1.Total_Points > 57.5 # ### End.
gymnast	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # gymnast ( Gymnast_ID, Floor_Exercise_Points, Pommel_Horse_Points, Rings_Points, Vault_Points, Parallel_Bars_Points, Horizontal_Bar_Points, Total_Points ) # people ( People_ID, Name, Age, Height, Hometown ) # # gymnast.Gymnast_ID can be joined with people.People_ID # ### Question: # # Give the different hometowns of gymnasts that have a total point score of above 57.5. # ### SQL: # # SELECT DISTINCT T2.Hometown FROM gymnast AS T1 JOIN people AS T2 ON T1.Gymnast_ID = T2.People_ID WHERE T1.Total_Points > 57.5 # ### End.
gymnast	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # gymnast ( Gymnast_ID, Floor_Exercise_Points, Pommel_Horse_Points, Rings_Points, Vault_Points, Parallel_Bars_Points, Horizontal_Bar_Points, Total_Points ) # people ( People_ID, Name, Age, Height, Hometown ) # # gymnast.Gymnast_ID can be joined with people.People_ID # ### Question: # # What are the hometowns of gymnasts and the corresponding number of gymnasts? # ### SQL: # # SELECT T2.Hometown , COUNT(*) FROM gymnast AS T1 JOIN people AS T2 ON T1.Gymnast_ID = T2.People_ID GROUP BY T2.Hometown # ### End.
gymnast	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # gymnast ( Gymnast_ID, Floor_Exercise_Points, Pommel_Horse_Points, Rings_Points, Vault_Points, Parallel_Bars_Points, Horizontal_Bar_Points, Total_Points ) # people ( People_ID, Name, Age, Height, Hometown ) # # gymnast.Gymnast_ID can be joined with people.People_ID # ### Question: # # How many gymnasts are from each hometown? # ### SQL: # # SELECT T2.Hometown , COUNT(*) FROM gymnast AS T1 JOIN people AS T2 ON T1.Gymnast_ID = T2.People_ID GROUP BY T2.Hometown # ### End.
gymnast	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # gymnast ( Gymnast_ID, Floor_Exercise_Points, Pommel_Horse_Points, Rings_Points, Vault_Points, Parallel_Bars_Points, Horizontal_Bar_Points, Total_Points ) # people ( People_ID, Name, Age, Height, Hometown ) # # gymnast.Gymnast_ID can be joined with people.People_ID # ### Question: # # What is the most common hometown of gymnasts? # ### SQL: # # SELECT T2.Hometown FROM gymnast AS T1 JOIN people AS T2 ON T1.Gymnast_ID = T2.People_ID GROUP BY T2.Hometown ORDER BY COUNT(*) DESC LIMIT 1 # ### End.
gymnast	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # gymnast ( Gymnast_ID, Floor_Exercise_Points, Pommel_Horse_Points, Rings_Points, Vault_Points, Parallel_Bars_Points, Horizontal_Bar_Points, Total_Points ) # people ( People_ID, Name, Age, Height, Hometown ) # # gymnast.Gymnast_ID can be joined with people.People_ID # ### Question: # # Return the hometown that is most common among gymnasts. # ### SQL: # # SELECT T2.Hometown FROM gymnast AS T1 JOIN people AS T2 ON T1.Gymnast_ID = T2.People_ID GROUP BY T2.Hometown ORDER BY COUNT(*) DESC LIMIT 1 # ### End.
gymnast	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # gymnast ( Gymnast_ID, Floor_Exercise_Points, Pommel_Horse_Points, Rings_Points, Vault_Points, Parallel_Bars_Points, Horizontal_Bar_Points, Total_Points ) # people ( People_ID, Name, Age, Height, Hometown ) # # gymnast.Gymnast_ID can be joined with people.People_ID # ### Question: # # What are the hometowns that are shared by at least two gymnasts? # ### SQL: # # SELECT T2.Hometown FROM gymnast AS T1 JOIN people AS T2 ON T1.Gymnast_ID = T2.People_ID GROUP BY T2.Hometown HAVING COUNT(*) >= 2 # ### End.
gymnast	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # gymnast ( Gymnast_ID, Floor_Exercise_Points, Pommel_Horse_Points, Rings_Points, Vault_Points, Parallel_Bars_Points, Horizontal_Bar_Points, Total_Points ) # people ( People_ID, Name, Age, Height, Hometown ) # # gymnast.Gymnast_ID can be joined with people.People_ID # ### Question: # # Give the hometowns from which two or more gymnasts are from. # ### SQL: # # SELECT T2.Hometown FROM gymnast AS T1 JOIN people AS T2 ON T1.Gymnast_ID = T2.People_ID GROUP BY T2.Hometown HAVING COUNT(*) >= 2 # ### End.
gymnast	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # gymnast ( Gymnast_ID, Floor_Exercise_Points, Pommel_Horse_Points, Rings_Points, Vault_Points, Parallel_Bars_Points, Horizontal_Bar_Points, Total_Points ) # people ( People_ID, Name, Age, Height, Hometown ) # # gymnast.Gymnast_ID can be joined with people.People_ID # ### Question: # # List the names of gymnasts in ascending order by their heights. # ### SQL: # # SELECT T2.Name FROM gymnast AS T1 JOIN people AS T2 ON T1.Gymnast_ID = T2.People_ID ORDER BY T2.Height ASC # ### End.
gymnast	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # gymnast ( Gymnast_ID, Floor_Exercise_Points, Pommel_Horse_Points, Rings_Points, Vault_Points, Parallel_Bars_Points, Horizontal_Bar_Points, Total_Points ) # people ( People_ID, Name, Age, Height, Hometown ) # # gymnast.Gymnast_ID can be joined with people.People_ID # ### Question: # # What are the names of gymnasts, ordered by their heights ascending? # ### SQL: # # SELECT T2.Name FROM gymnast AS T1 JOIN people AS T2 ON T1.Gymnast_ID = T2.People_ID ORDER BY T2.Height ASC # ### End.
gymnast	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # gymnast ( Gymnast_ID, Floor_Exercise_Points, Pommel_Horse_Points, Rings_Points, Vault_Points, Parallel_Bars_Points, Horizontal_Bar_Points, Total_Points ) # people ( People_ID, Name, Age, Height, Hometown ) # # gymnast.Gymnast_ID can be joined with people.People_ID # ### Question: # # List the distinct hometowns that are not associated with any gymnast. # ### SQL: # # SELECT DISTINCT Hometown FROM people EXCEPT SELECT DISTINCT T2.Hometown FROM gymnast AS T1 JOIN people AS T2 ON T1.Gymnast_ID = T2.People_ID # ### End.
gymnast	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # gymnast ( Gymnast_ID, Floor_Exercise_Points, Pommel_Horse_Points, Rings_Points, Vault_Points, Parallel_Bars_Points, Horizontal_Bar_Points, Total_Points ) # people ( People_ID, Name, Age, Height, Hometown ) # # gymnast.Gymnast_ID can be joined with people.People_ID # ### Question: # # From which hometowns did no gymnasts come from? # ### SQL: # # SELECT DISTINCT Hometown FROM people EXCEPT SELECT DISTINCT T2.Hometown FROM gymnast AS T1 JOIN people AS T2 ON T1.Gymnast_ID = T2.People_ID # ### End.
gymnast	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # gymnast ( Gymnast_ID, Floor_Exercise_Points, Pommel_Horse_Points, Rings_Points, Vault_Points, Parallel_Bars_Points, Horizontal_Bar_Points, Total_Points ) # people ( People_ID, Name, Age, Height, Hometown ) # # gymnast.Gymnast_ID can be joined with people.People_ID # ### Question: # # Show the hometowns shared by people older than 23 and younger than 20. # ### SQL: # # SELECT Hometown FROM people WHERE Age > 23 INTERSECT SELECT Hometown FROM people WHERE Age < 20 # ### End.
gymnast	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # gymnast ( Gymnast_ID, Floor_Exercise_Points, Pommel_Horse_Points, Rings_Points, Vault_Points, Parallel_Bars_Points, Horizontal_Bar_Points, Total_Points ) # people ( People_ID, Name, Age, Height, Hometown ) # # gymnast.Gymnast_ID can be joined with people.People_ID # ### Question: # # From which hometowns did both people older than 23 and younger than 20 come from? # ### SQL: # # SELECT Hometown FROM people WHERE Age > 23 INTERSECT SELECT Hometown FROM people WHERE Age < 20 # ### End.
gymnast	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # gymnast ( Gymnast_ID, Floor_Exercise_Points, Pommel_Horse_Points, Rings_Points, Vault_Points, Parallel_Bars_Points, Horizontal_Bar_Points, Total_Points ) # people ( People_ID, Name, Age, Height, Hometown ) # # gymnast.Gymnast_ID can be joined with people.People_ID # ### Question: # # How many distinct hometowns did these people have? # ### SQL: # # SELECT count(DISTINCT Hometown) FROM people # ### End.
gymnast	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # gymnast ( Gymnast_ID, Floor_Exercise_Points, Pommel_Horse_Points, Rings_Points, Vault_Points, Parallel_Bars_Points, Horizontal_Bar_Points, Total_Points ) # people ( People_ID, Name, Age, Height, Hometown ) # # gymnast.Gymnast_ID can be joined with people.People_ID # ### Question: # # Count the number of different hometowns of these people. # ### SQL: # # SELECT count(DISTINCT Hometown) FROM people # ### End.
gymnast	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # gymnast ( Gymnast_ID, Floor_Exercise_Points, Pommel_Horse_Points, Rings_Points, Vault_Points, Parallel_Bars_Points, Horizontal_Bar_Points, Total_Points ) # people ( People_ID, Name, Age, Height, Hometown ) # # gymnast.Gymnast_ID can be joined with people.People_ID # ### Question: # # Show the ages of gymnasts in descending order of total points. # ### SQL: # # SELECT T2.Age FROM gymnast AS T1 JOIN people AS T2 ON T1.Gymnast_ID = T2.People_ID ORDER BY T1.Total_Points DESC # ### End.
gymnast	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # gymnast ( Gymnast_ID, Floor_Exercise_Points, Pommel_Horse_Points, Rings_Points, Vault_Points, Parallel_Bars_Points, Horizontal_Bar_Points, Total_Points ) # people ( People_ID, Name, Age, Height, Hometown ) # # gymnast.Gymnast_ID can be joined with people.People_ID # ### Question: # # What are the ages of the gymnasts, ordered descending by their total points? # ### SQL: # # SELECT T2.Age FROM gymnast AS T1 JOIN people AS T2 ON T1.Gymnast_ID = T2.People_ID ORDER BY T1.Total_Points DESC # ### End.
small_bank_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # ACCOUNTS ( custid, name ) # SAVINGS ( custid, balance ) # CHECKING ( custid, balance ) # # SAVINGS.custid can be joined with ACCOUNTS.custid # CHECKING.custid can be joined with ACCOUNTS.custid # ### Question: # # Find the total savings balance of all accounts except the account with name ‘Brown’. # ### SQL: # # SELECT sum(T2.balance) FROM accounts AS T1 JOIN savings AS T2 ON T1.custid = T2.custid WHERE T1.name != 'Brown' # ### End.
small_bank_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # ACCOUNTS ( custid, name ) # SAVINGS ( custid, balance ) # CHECKING ( custid, balance ) # # SAVINGS.custid can be joined with ACCOUNTS.custid # CHECKING.custid can be joined with ACCOUNTS.custid # ### Question: # # What is the total balance of savings accounts not belonging to someone with the name Brown? # ### SQL: # # SELECT sum(T2.balance) FROM accounts AS T1 JOIN savings AS T2 ON T1.custid = T2.custid WHERE T1.name != 'Brown' # ### End.
small_bank_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # ACCOUNTS ( custid, name ) # SAVINGS ( custid, balance ) # CHECKING ( custid, balance ) # # SAVINGS.custid can be joined with ACCOUNTS.custid # CHECKING.custid can be joined with ACCOUNTS.custid # ### Question: # # How many accounts are there in total? # ### SQL: # # SELECT count(*) FROM accounts # ### End.
small_bank_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # ACCOUNTS ( custid, name ) # SAVINGS ( custid, balance ) # CHECKING ( custid, balance ) # # SAVINGS.custid can be joined with ACCOUNTS.custid # CHECKING.custid can be joined with ACCOUNTS.custid # ### Question: # # Count the number of accounts. # ### SQL: # # SELECT count(*) FROM accounts # ### End.
small_bank_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # ACCOUNTS ( custid, name ) # SAVINGS ( custid, balance ) # CHECKING ( custid, balance ) # # SAVINGS.custid can be joined with ACCOUNTS.custid # CHECKING.custid can be joined with ACCOUNTS.custid # ### Question: # # What is the total checking balance in all accounts? # ### SQL: # # SELECT sum(balance) FROM checking # ### End.
small_bank_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # ACCOUNTS ( custid, name ) # SAVINGS ( custid, balance ) # CHECKING ( custid, balance ) # # SAVINGS.custid can be joined with ACCOUNTS.custid # CHECKING.custid can be joined with ACCOUNTS.custid # ### Question: # # Find the total balance across checking accounts. # ### SQL: # # SELECT sum(balance) FROM checking # ### End.
small_bank_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # ACCOUNTS ( custid, name ) # SAVINGS ( custid, balance ) # CHECKING ( custid, balance ) # # SAVINGS.custid can be joined with ACCOUNTS.custid # CHECKING.custid can be joined with ACCOUNTS.custid # ### Question: # # Find the average checking balance. # ### SQL: # # SELECT avg(balance) FROM checking # ### End.
small_bank_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # ACCOUNTS ( custid, name ) # SAVINGS ( custid, balance ) # CHECKING ( custid, balance ) # # SAVINGS.custid can be joined with ACCOUNTS.custid # CHECKING.custid can be joined with ACCOUNTS.custid # ### Question: # # What is the average balance in checking accounts? # ### SQL: # # SELECT avg(balance) FROM checking # ### End.
small_bank_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # ACCOUNTS ( custid, name ) # SAVINGS ( custid, balance ) # CHECKING ( custid, balance ) # # SAVINGS.custid can be joined with ACCOUNTS.custid # CHECKING.custid can be joined with ACCOUNTS.custid # ### Question: # # How many accounts have a savings balance above the average savings balance? # ### SQL: # # SELECT count(*) FROM savings WHERE balance > (SELECT avg(balance) FROM savings) # ### End.
small_bank_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # ACCOUNTS ( custid, name ) # SAVINGS ( custid, balance ) # CHECKING ( custid, balance ) # # SAVINGS.custid can be joined with ACCOUNTS.custid # CHECKING.custid can be joined with ACCOUNTS.custid # ### Question: # # Find the number of accounts with a savings balance that is higher than the average savings balance. # ### SQL: # # SELECT count(*) FROM savings WHERE balance > (SELECT avg(balance) FROM savings) # ### End.
small_bank_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # ACCOUNTS ( custid, name ) # SAVINGS ( custid, balance ) # CHECKING ( custid, balance ) # # SAVINGS.custid can be joined with ACCOUNTS.custid # CHECKING.custid can be joined with ACCOUNTS.custid # ### Question: # # Find the name and id of accounts whose checking balance is below the maximum checking balance. # ### SQL: # # SELECT T1.custid , T1.name FROM accounts AS T1 JOIN checking AS T2 ON T1.custid = T2.custid WHERE T2.balance < (SELECT max(balance) FROM checking) # ### End.
small_bank_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # ACCOUNTS ( custid, name ) # SAVINGS ( custid, balance ) # CHECKING ( custid, balance ) # # SAVINGS.custid can be joined with ACCOUNTS.custid # CHECKING.custid can be joined with ACCOUNTS.custid # ### Question: # # What are the customer id and name corresponding to accounts with a checking balance less than the largest checking balance? # ### SQL: # # SELECT T1.custid , T1.name FROM accounts AS T1 JOIN checking AS T2 ON T1.custid = T2.custid WHERE T2.balance < (SELECT max(balance) FROM checking) # ### End.
small_bank_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # ACCOUNTS ( custid, name ) # SAVINGS ( custid, balance ) # CHECKING ( custid, balance ) # # SAVINGS.custid can be joined with ACCOUNTS.custid # CHECKING.custid can be joined with ACCOUNTS.custid # ### Question: # # What is the checking balance of the account whose owner’s name contains the substring ‘ee’? # ### SQL: # # SELECT T2.balance FROM accounts AS T1 JOIN checking AS T2 ON T1.custid = T2.custid WHERE T1.name LIKE '%ee%' # ### End.
small_bank_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # ACCOUNTS ( custid, name ) # SAVINGS ( custid, balance ) # CHECKING ( custid, balance ) # # SAVINGS.custid can be joined with ACCOUNTS.custid # CHECKING.custid can be joined with ACCOUNTS.custid # ### Question: # # Find the balance of the checking account belonging to an owner whose name contains 'ee'. # ### SQL: # # SELECT T2.balance FROM accounts AS T1 JOIN checking AS T2 ON T1.custid = T2.custid WHERE T1.name LIKE '%ee%' # ### End.
small_bank_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # ACCOUNTS ( custid, name ) # SAVINGS ( custid, balance ) # CHECKING ( custid, balance ) # # SAVINGS.custid can be joined with ACCOUNTS.custid # CHECKING.custid can be joined with ACCOUNTS.custid # ### Question: # # Find the checking balance and saving balance in the Brown’s account. # ### SQL: # # SELECT T2.balance , T3.balance FROM accounts AS T1 JOIN checking AS T2 ON T1.custid = T2.custid JOIN savings AS T3 ON T1.custid = T3.custid WHERE T1.name = 'Brown' # ### End.
small_bank_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # ACCOUNTS ( custid, name ) # SAVINGS ( custid, balance ) # CHECKING ( custid, balance ) # # SAVINGS.custid can be joined with ACCOUNTS.custid # CHECKING.custid can be joined with ACCOUNTS.custid # ### Question: # # What are the checking and savings balances in accounts belonging to Brown? # ### SQL: # # SELECT T2.balance , T3.balance FROM accounts AS T1 JOIN checking AS T2 ON T1.custid = T2.custid JOIN savings AS T3 ON T1.custid = T3.custid WHERE T1.name = 'Brown' # ### End.
small_bank_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # ACCOUNTS ( custid, name ) # SAVINGS ( custid, balance ) # CHECKING ( custid, balance ) # # SAVINGS.custid can be joined with ACCOUNTS.custid # CHECKING.custid can be joined with ACCOUNTS.custid # ### Question: # # Find the names of accounts whose checking balance is above the average checking balance, but savings balance is below the average savings balance. # ### SQL: # # SELECT T1.name FROM accounts AS T1 JOIN checking AS T2 ON T1.custid = T2.custid WHERE T2.balance > (SELECT avg(balance) FROM checking) INTERSECT SELECT T1.name FROM accounts AS T1 JOIN savings AS T2 ON T1.custid = T2.custid WHERE T2.balance < (SELECT avg(balance) FROM savings) # ### End.
small_bank_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # ACCOUNTS ( custid, name ) # SAVINGS ( custid, balance ) # CHECKING ( custid, balance ) # # SAVINGS.custid can be joined with ACCOUNTS.custid # CHECKING.custid can be joined with ACCOUNTS.custid # ### Question: # # What are the names of accounts with checking balances greater than the average checking balance and savings balances below the average savings balance? # ### SQL: # # SELECT T1.name FROM accounts AS T1 JOIN checking AS T2 ON T1.custid = T2.custid WHERE T2.balance > (SELECT avg(balance) FROM checking) INTERSECT SELECT T1.name FROM accounts AS T1 JOIN savings AS T2 ON T1.custid = T2.custid WHERE T2.balance < (SELECT avg(balance) FROM savings) # ### End.
small_bank_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # ACCOUNTS ( custid, name ) # SAVINGS ( custid, balance ) # CHECKING ( custid, balance ) # # SAVINGS.custid can be joined with ACCOUNTS.custid # CHECKING.custid can be joined with ACCOUNTS.custid # ### Question: # # Find the checking balance of the accounts whose savings balance is higher than the average savings balance. # ### SQL: # # SELECT T2.balance FROM accounts AS T1 JOIN checking AS T2 ON T1.custid = T2.custid WHERE T1.name IN (SELECT T1.name FROM accounts AS T1 JOIN savings AS T2 ON T1.custid = T2.custid WHERE T2.balance > (SELECT avg(balance) FROM savings)) # ### End.
small_bank_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # ACCOUNTS ( custid, name ) # SAVINGS ( custid, balance ) # CHECKING ( custid, balance ) # # SAVINGS.custid can be joined with ACCOUNTS.custid # CHECKING.custid can be joined with ACCOUNTS.custid # ### Question: # # What are the balances of checking accounts belonging to people with savings balances greater than the average savings balance? # ### SQL: # # SELECT T2.balance FROM accounts AS T1 JOIN checking AS T2 ON T1.custid = T2.custid WHERE T1.name IN (SELECT T1.name FROM accounts AS T1 JOIN savings AS T2 ON T1.custid = T2.custid WHERE T2.balance > (SELECT avg(balance) FROM savings)) # ### End.
small_bank_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # ACCOUNTS ( custid, name ) # SAVINGS ( custid, balance ) # CHECKING ( custid, balance ) # # SAVINGS.custid can be joined with ACCOUNTS.custid # CHECKING.custid can be joined with ACCOUNTS.custid # ### Question: # # List all customers’ names in the alphabetical order. # ### SQL: # # SELECT name FROM accounts ORDER BY name # ### End.
small_bank_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # ACCOUNTS ( custid, name ) # SAVINGS ( custid, balance ) # CHECKING ( custid, balance ) # # SAVINGS.custid can be joined with ACCOUNTS.custid # CHECKING.custid can be joined with ACCOUNTS.custid # ### Question: # # What are the names of all the customers in alphabetical order? # ### SQL: # # SELECT name FROM accounts ORDER BY name # ### End.
small_bank_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # ACCOUNTS ( custid, name ) # SAVINGS ( custid, balance ) # CHECKING ( custid, balance ) # # SAVINGS.custid can be joined with ACCOUNTS.custid # CHECKING.custid can be joined with ACCOUNTS.custid # ### Question: # # Find the name of account that has the lowest total checking and saving balance. # ### SQL: # # SELECT T1.name FROM accounts AS T1 JOIN checking AS T2 ON T1.custid = T2.custid JOIN savings AS T3 ON T1.custid = T3.custid ORDER BY T2.balance + T3.balance LIMIT 1 # ### End.
small_bank_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # ACCOUNTS ( custid, name ) # SAVINGS ( custid, balance ) # CHECKING ( custid, balance ) # # SAVINGS.custid can be joined with ACCOUNTS.custid # CHECKING.custid can be joined with ACCOUNTS.custid # ### Question: # # What is the name corresponding to the accoung with the lowest sum of checking and savings balances? # ### SQL: # # SELECT T1.name FROM accounts AS T1 JOIN checking AS T2 ON T1.custid = T2.custid JOIN savings AS T3 ON T1.custid = T3.custid ORDER BY T2.balance + T3.balance LIMIT 1 # ### End.

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