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Encontre o nome dos instrutores que estão aconselhando mais de um aluno.
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CREATE TABLE advisor (i_id VARCHAR); CREATE TABLE instructor (name VARCHAR, id VARCHAR)
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SELECT T1.name FROM instructor AS T1 JOIN advisor AS T2 ON T1.id = T2.i_id GROUP BY T2.i_id HAVING COUNT(*) > 1
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Encontrar o nome dos alunos que têm mais de um conselheiro?
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CREATE TABLE student (name VARCHAR, id VARCHAR); CREATE TABLE advisor (s_id VARCHAR)
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SELECT T1.name FROM student AS T1 JOIN advisor AS T2 ON T1.id = T2.s_id GROUP BY T2.s_id HAVING COUNT(*) > 1
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Encontre o número de salas com capacidade superior a 50 para cada prédio.
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CREATE TABLE classroom (building VARCHAR, capacity INTEGER)
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SELECT COUNT(*), building FROM classroom WHERE capacity > 50 GROUP BY building
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Encontre a capacidade máxima e média entre os quartos de cada edifício.
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CREATE TABLE classroom (building VARCHAR, capacity INTEGER)
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SELECT MAX(capacity), AVG(capacity), building FROM classroom GROUP BY building
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Encontre o título do curso que é oferecido por mais de um departamento.
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CREATE TABLE course (title VARCHAR)
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SELECT title FROM course GROUP BY title HAVING COUNT(*) > 1
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Encontre o total de créditos de cursos oferecidos por diferentes departamentos.
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CREATE TABLE course (dept_name VARCHAR, credits INTEGER)
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SELECT SUM(credits), dept_name FROM course GROUP BY dept_name
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Determine o salário mínimo para os departamentos cujo salário médio é superior ao salário médio de todos os professores.
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CREATE TABLE instructor (dept_name VARCHAR, salary INTEGER)
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SELECT MIN(salary), dept_name FROM instructor GROUP BY dept_name HAVING AVG(salary) > (SELECT AVG(salary) FROM instructor)
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Encontre o número de cursos oferecidos em cada semestre e ano.
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CREATE TABLE SECTION (semester VARCHAR, YEAR VARCHAR)
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SELECT COUNT(*), semester, YEAR FROM SECTION GROUP BY semester, YEAR
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Encontre o ano que oferece o maior número de cursos.
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CREATE TABLE SECTION (YEAR VARCHAR)
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SELECT YEAR FROM SECTION GROUP BY YEAR ORDER BY COUNT(*) DESC LIMIT 1
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Encontre o ano e o semestre em que o maior número de cursos é oferecido.
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CREATE TABLE SECTION (semester VARCHAR, YEAR VARCHAR)
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SELECT semester, YEAR FROM SECTION GROUP BY semester, YEAR ORDER BY COUNT(*) DESC LIMIT 1
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Encontrar o nome do departamento tem o maior número de estudantes?
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CREATE TABLE student (dept_name VARCHAR)
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SELECT dept_name FROM student GROUP BY dept_name ORDER BY COUNT(*) DESC LIMIT 1
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Encontre o número total de alunos em cada departamento.
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CREATE TABLE student (dept_name VARCHAR)
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SELECT COUNT(*), dept_name FROM student GROUP BY dept_name
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Encontre o semestre e o ano que tem o menor número de alunos que fazem qualquer aula.
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CREATE TABLE takes (semester VARCHAR, YEAR VARCHAR)
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SELECT semester, YEAR FROM takes GROUP BY semester, YEAR ORDER BY COUNT(*) LIMIT 1
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Qual é o id do instrutor que aconselha todos os alunos do departamento de História?
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CREATE TABLE advisor (s_id VARCHAR); CREATE TABLE student (id VARCHAR, dept_name VARCHAR)
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SELECT i_id FROM advisor AS T1 JOIN student AS T2 ON T1.s_id = T2.id WHERE T2.dept_name = 'History'
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Encontrar o nome e o salário dos instrutores que são conselheiros de qualquer aluno do departamento de História?
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CREATE TABLE instructor (name VARCHAR, salary VARCHAR, id VARCHAR); CREATE TABLE advisor (i_id VARCHAR, s_id VARCHAR); CREATE TABLE student (id VARCHAR, dept_name VARCHAR)
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SELECT T2.name, T2.salary FROM advisor AS T1 JOIN instructor AS T2 ON T1.i_id = T2.id JOIN student AS T3 ON T1.s_id = T3.id WHERE T3.dept_name = 'History'
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Encontrar o id dos cursos que não têm qualquer pré-requisito?
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CREATE TABLE prereq (course_id VARCHAR); CREATE TABLE course (course_id VARCHAR)
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SELECT course_id FROM course EXCEPT SELECT course_id FROM prereq
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Qual é o título da aula de pré-requisitos do curso de Finanças Internacionais?
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CREATE TABLE course (title VARCHAR, course_id VARCHAR); CREATE TABLE prereq (prereq_id VARCHAR, course_id VARCHAR); CREATE TABLE course (course_id VARCHAR, title VARCHAR)
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SELECT title FROM course WHERE course_id IN (SELECT T1.prereq_id FROM prereq AS T1 JOIN course AS T2 ON T1.course_id = T2.course_id WHERE T2.title = 'International Finance')
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Encontre o título do curso cujo pré-requisito é o curso Geometria Diferencial.
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CREATE TABLE prereq (course_id VARCHAR, prereq_id VARCHAR); CREATE TABLE course (title VARCHAR, course_id VARCHAR); CREATE TABLE course (course_id VARCHAR, title VARCHAR)
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SELECT title FROM course WHERE course_id IN (SELECT T1.course_id FROM prereq AS T1 JOIN course AS T2 ON T1.prereq_id = T2.course_id WHERE T2.title = 'Differential Geometry')
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Encontre os nomes dos estudantes que fizeram qualquer curso no semestre de outono do ano 2003.
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CREATE TABLE student (name VARCHAR, id VARCHAR, semester VARCHAR, YEAR VARCHAR); CREATE TABLE takes (name VARCHAR, id VARCHAR, semester VARCHAR, YEAR VARCHAR)
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SELECT name FROM student WHERE id IN (SELECT id FROM takes WHERE semester = 'Fall' AND YEAR = 2003)
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Qual é o título do curso que foi oferecido no edifício Chandler durante o semestre de outono no ano de 2010?
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CREATE TABLE course (title VARCHAR, course_id VARCHAR); CREATE TABLE SECTION (course_id VARCHAR)
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SELECT T1.title FROM course AS T1 JOIN SECTION AS T2 ON T1.course_id = T2.course_id WHERE building = 'Chandler' AND semester = 'Fall' AND YEAR = 2010
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Encontre o nome dos instrutores que ensinaram C programação curso antes.
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CREATE TABLE teaches (id VARCHAR, course_id VARCHAR); CREATE TABLE course (course_id VARCHAR, title VARCHAR); CREATE TABLE instructor (name VARCHAR, id VARCHAR)
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SELECT T1.name FROM instructor AS T1 JOIN teaches AS T2 ON T1.id = T2.id JOIN course AS T3 ON T2.course_id = T3.course_id WHERE T3.title = 'C Programming'
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Encontre o nome e o salário dos instrutores que são conselheiros dos alunos do departamento de Matemática.
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CREATE TABLE instructor (name VARCHAR, salary VARCHAR, id VARCHAR); CREATE TABLE advisor (i_id VARCHAR, s_id VARCHAR); CREATE TABLE student (id VARCHAR, dept_name VARCHAR)
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SELECT T2.name, T2.salary FROM advisor AS T1 JOIN instructor AS T2 ON T1.i_id = T2.id JOIN student AS T3 ON T1.s_id = T3.id WHERE T3.dept_name = 'Math'
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Encontre o nome dos instrutores que são conselheiros dos alunos do departamento de Matemática e classifique os resultados por crédito total dos alunos.
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CREATE TABLE student (id VARCHAR, dept_name VARCHAR, tot_cred VARCHAR); CREATE TABLE advisor (i_id VARCHAR, s_id VARCHAR); CREATE TABLE instructor (name VARCHAR, id VARCHAR)
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SELECT T2.name FROM advisor AS T1 JOIN instructor AS T2 ON T1.i_id = T2.id JOIN student AS T3 ON T1.s_id = T3.id WHERE T3.dept_name = 'Math' ORDER BY T3.tot_cred
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Qual é o título do curso do pré-requisito do curso de Computação Móvel?
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CREATE TABLE course (title VARCHAR, course_id VARCHAR); CREATE TABLE prereq (prereq_id VARCHAR, course_id VARCHAR); CREATE TABLE course (course_id VARCHAR, title VARCHAR)
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SELECT title FROM course WHERE course_id IN (SELECT T1.prereq_id FROM prereq AS T1 JOIN course AS T2 ON T1.course_id = T2.course_id WHERE T2.title = 'Mobile Computing')
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Encontre o nome do instrutor que é o conselheiro do aluno que tem o maior número de créditos totais.
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CREATE TABLE student (id VARCHAR, tot_cred VARCHAR); CREATE TABLE advisor (i_id VARCHAR, s_id VARCHAR); CREATE TABLE instructor (name VARCHAR, id VARCHAR)
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SELECT T2.name FROM advisor AS T1 JOIN instructor AS T2 ON T1.i_id = T2.id JOIN student AS T3 ON T1.s_id = T3.id ORDER BY T3.tot_cred DESC LIMIT 1
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Encontrar o nome de instrutores que não ensinaram nenhum curso?
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CREATE TABLE teaches (name VARCHAR, id VARCHAR); CREATE TABLE instructor (name VARCHAR, id VARCHAR)
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SELECT name FROM instructor WHERE NOT id IN (SELECT id FROM teaches)
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Encontrar a identidade dos instrutores que não ensinaram nenhum curso?
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CREATE TABLE teaches (id VARCHAR); CREATE TABLE instructor (id VARCHAR)
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SELECT id FROM instructor EXCEPT SELECT id FROM teaches
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Encontre os nomes dos instrutores que não fizeram nenhum curso em nenhum semestre de primavera.
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CREATE TABLE teaches (name VARCHAR, id VARCHAR, semester VARCHAR); CREATE TABLE instructor (name VARCHAR, id VARCHAR, semester VARCHAR)
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SELECT name FROM instructor WHERE NOT id IN (SELECT id FROM teaches WHERE semester = 'Spring')
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Encontre o nome do departamento que tem o salário médio mais alto de professores.
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CREATE TABLE instructor (dept_name VARCHAR, salary INTEGER)
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SELECT dept_name FROM instructor GROUP BY dept_name ORDER BY AVG(salary) DESC LIMIT 1
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Encontre o número e o salário médio de todos os instrutores que estão no departamento com o maior orçamento.
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CREATE TABLE department (dept_name VARCHAR, budget VARCHAR); CREATE TABLE instructor (salary INTEGER, dept_name VARCHAR)
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SELECT AVG(T1.salary), COUNT(*) FROM instructor AS T1 JOIN department AS T2 ON T1.dept_name = T2.dept_name ORDER BY T2.budget DESC LIMIT 1
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Qual é o título e os créditos do curso que é ensinado na sala de aula mais grande (com a maior capacidade)?
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CREATE TABLE SECTION (course_id VARCHAR, building VARCHAR, room_number VARCHAR); CREATE TABLE course (title VARCHAR, credits VARCHAR, course_id VARCHAR); CREATE TABLE classroom (capacity INTEGER, building VARCHAR, room_number VARCHAR); CREATE TABLE classroom (capacity INTEGER)
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SELECT T3.title, T3.credits FROM classroom AS T1 JOIN SECTION AS T2 ON T1.building = T2.building AND T1.room_number = T2.room_number JOIN course AS T3 ON T2.course_id = T3.course_id WHERE T1.capacity = (SELECT MAX(capacity) FROM classroom)
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Encontre o nome dos estudantes que não fizeram nenhum curso do departamento de Biologia.
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CREATE TABLE student (name VARCHAR, id VARCHAR); CREATE TABLE course (course_id VARCHAR, dept_name VARCHAR); CREATE TABLE takes (id VARCHAR, course_id VARCHAR)
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SELECT name FROM student WHERE NOT id IN (SELECT T1.id FROM takes AS T1 JOIN course AS T2 ON T1.course_id = T2.course_id WHERE T2.dept_name = 'Biology')
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Encontre o número total de alunos e o número total de instrutores para cada departamento.
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CREATE TABLE department (dept_name VARCHAR); CREATE TABLE student (id VARCHAR, dept_name VARCHAR); CREATE TABLE instructor (dept_name VARCHAR, id VARCHAR)
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SELECT COUNT(DISTINCT T2.id), COUNT(DISTINCT T3.id), T3.dept_name FROM department AS T1 JOIN student AS T2 ON T1.dept_name = T2.dept_name JOIN instructor AS T3 ON T1.dept_name = T3.dept_name GROUP BY T3.dept_name
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Encontre o nome dos estudantes que fizeram o curso de pré-requisito do curso com o título Finanças Internacionais.
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CREATE TABLE student (name VARCHAR, id VARCHAR); CREATE TABLE course (course_id VARCHAR, title VARCHAR); CREATE TABLE prereq (prereq_id VARCHAR, course_id VARCHAR); CREATE TABLE takes (id VARCHAR, course_id VARCHAR)
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SELECT T1.name FROM student AS T1 JOIN takes AS T2 ON T1.id = T2.id WHERE T2.course_id IN (SELECT T4.prereq_id FROM course AS T3 JOIN prereq AS T4 ON T3.course_id = T4.course_id WHERE T3.title = 'International Finance')
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Encontre o nome e o salário dos instrutores cujo salário é inferior ao salário médio dos instrutores do departamento de Física.
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CREATE TABLE instructor (name VARCHAR, salary INTEGER, dept_name VARCHAR)
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SELECT name, salary FROM instructor WHERE salary < (SELECT AVG(salary) FROM instructor WHERE dept_name = 'Physics')
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Encontre o nome de estudantes que fizeram algum curso oferecido pelo departamento de Estatística.
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CREATE TABLE student (name VARCHAR, id VARCHAR); CREATE TABLE takes (course_id VARCHAR, id VARCHAR); CREATE TABLE course (course_id VARCHAR, dept_name VARCHAR)
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SELECT T3.name FROM course AS T1 JOIN takes AS T2 ON T1.course_id = T2.course_id JOIN student AS T3 ON T2.id = T3.id WHERE T1.dept_name = 'Statistics'
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Encontre o prédio, número de quarto, semestre e ano de todos os cursos oferecidos pelo departamento de Psicologia ordenados por títulos de cursos.
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CREATE TABLE SECTION (building VARCHAR, room_number VARCHAR, semester VARCHAR, year VARCHAR, course_id VARCHAR); CREATE TABLE course (course_id VARCHAR, dept_name VARCHAR, title VARCHAR)
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SELECT T2.building, T2.room_number, T2.semester, T2.year FROM course AS T1 JOIN SECTION AS T2 ON T1.course_id = T2.course_id WHERE T1.dept_name = 'Psychology' ORDER BY T1.title
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Encontre os nomes de todos os instrutores no departamento de ciência da computação
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CREATE TABLE instructor (name VARCHAR, dept_name VARCHAR)
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SELECT name FROM instructor WHERE dept_name = 'Comp. Sci.'
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Encontre os nomes de todos os instrutores do departamento de ciências da empresa com salário > 80000.
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CREATE TABLE instructor (name VARCHAR, dept_name VARCHAR, salary VARCHAR)
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SELECT name FROM instructor WHERE dept_name = 'Comp. Sci.' AND salary > 80000
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Encontre os nomes de todos os instrutores que ensinaram algum curso e o curso_id.
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CREATE TABLE instructor (ID VARCHAR); CREATE TABLE teaches (ID VARCHAR)
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SELECT name, course_id FROM instructor AS T1 JOIN teaches AS T2 ON T1.ID = T2.ID
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Encontre os nomes de todos os instrutores do departamento de Arte que ensinaram algum curso e o curso_id.
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CREATE TABLE instructor (ID VARCHAR, dept_name VARCHAR); CREATE TABLE teaches (ID VARCHAR)
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SELECT name, course_id FROM instructor AS T1 JOIN teaches AS T2 ON T1.ID = T2.ID WHERE T1.dept_name = 'Art'
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Encontre os nomes de todos os instrutores cujo nome inclui a substring dar.
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CREATE TABLE instructor (name VARCHAR)
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SELECT name FROM instructor WHERE name LIKE '%dar%'
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Escreva em ordem alfabética os nomes de todos os instrutores.
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CREATE TABLE instructor (name VARCHAR)
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SELECT DISTINCT name FROM instructor ORDER BY name
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Encontre cursos que foram oferecidos no outono de 2009 ou na primavera de 2010.
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CREATE TABLE SECTION (course_id VARCHAR, semester VARCHAR, YEAR VARCHAR)
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SELECT course_id FROM SECTION WHERE semester = 'Fall' AND YEAR = 2009 UNION SELECT course_id FROM SECTION WHERE semester = 'Spring' AND YEAR = 2010
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Encontre cursos que foram realizados no Outono de 2009 e na Primavera de 2010.
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CREATE TABLE SECTION (course_id VARCHAR, semester VARCHAR, YEAR VARCHAR)
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SELECT course_id FROM SECTION WHERE semester = 'Fall' AND YEAR = 2009 INTERSECT SELECT course_id FROM SECTION WHERE semester = 'Spring' AND YEAR = 2010
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Encontre cursos que foram executados no outono de 2009 mas não na primavera de 2010.
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CREATE TABLE SECTION (course_id VARCHAR, semester VARCHAR, YEAR VARCHAR)
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SELECT course_id FROM SECTION WHERE semester = 'Fall' AND YEAR = 2009 EXCEPT SELECT course_id FROM SECTION WHERE semester = 'Spring' AND YEAR = 2010
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Encontre os salários de todos os instrutores distintos que são menores do salário maior.
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CREATE TABLE instructor (salary INTEGER)
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SELECT DISTINCT salary FROM instructor WHERE salary < (SELECT MAX(salary) FROM instructor)
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Encontre o número total de instrutores que ensinam um curso no semestre de primavera de 2010.
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CREATE TABLE teaches (ID VARCHAR, semester VARCHAR, YEAR VARCHAR)
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SELECT COUNT(DISTINCT ID) FROM teaches WHERE semester = 'Spring' AND YEAR = 2010
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Encontre os nomes e salários médios de todos os departamentos cujo salário médio é superior a 42000.
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CREATE TABLE instructor (dept_name VARCHAR, salary INTEGER)
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SELECT dept_name, AVG(salary) FROM instructor GROUP BY dept_name HAVING AVG(salary) > 42000
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Encontre nomes de instrutores com salário superior ao de algum (pelo menos um) instrutor no departamento de Biologia.
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CREATE TABLE instructor (name VARCHAR, salary INTEGER, dept_name VARCHAR)
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SELECT name FROM instructor WHERE salary > (SELECT MIN(salary) FROM instructor WHERE dept_name = 'Biology')
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Encontre os nomes de todos os instrutores cujo salário é maior que o salário de todos os instrutores do departamento de Biologia.
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CREATE TABLE instructor (name VARCHAR, salary INTEGER, dept_name VARCHAR)
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SELECT name FROM instructor WHERE salary > (SELECT MAX(salary) FROM instructor WHERE dept_name = 'Biology')
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Quantos debates há?
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CREATE TABLE debate (Id VARCHAR)
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SELECT COUNT(*) FROM debate
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Escreva os locais dos debates em ordem ascendente do número de pessoas presentes.
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CREATE TABLE debate (Venue VARCHAR, Num_of_Audience VARCHAR)
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SELECT Venue FROM debate ORDER BY Num_of_Audience
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Qual é a data e o local de cada debate?
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CREATE TABLE debate (Date VARCHAR, Venue VARCHAR)
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SELECT Date, Venue FROM debate
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Lista das datas dos debates com um número de audiências superior a 150
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CREATE TABLE debate (Date VARCHAR, Num_of_Audience INTEGER)
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SELECT Date FROM debate WHERE Num_of_Audience > 150
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Mostrar os nomes de pessoas com 35 ou 36 anos.
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CREATE TABLE people (Name VARCHAR, Age VARCHAR)
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SELECT Name FROM people WHERE Age = 35 OR Age = 36
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Qual é a festa dos mais jovens?
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CREATE TABLE people (Party VARCHAR, Age VARCHAR)
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SELECT Party FROM people ORDER BY Age LIMIT 1
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Mostrar diferentes grupos de pessoas, juntamente com o número de pessoas em cada grupo.
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CREATE TABLE people (Party VARCHAR)
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SELECT Party, COUNT(*) FROM people GROUP BY Party
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Mostra a festa que tem mais pessoas.
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CREATE TABLE people (Party VARCHAR)
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SELECT Party FROM people GROUP BY Party ORDER BY COUNT(*) DESC LIMIT 1
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Mostrar os diferentes locais de debate
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CREATE TABLE debate (Venue VARCHAR)
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SELECT DISTINCT Venue FROM debate
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Mostrar os nomes das pessoas, e datas e locais de debates que estão no lado afirmativo.
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CREATE TABLE people (Name VARCHAR, People_ID VARCHAR); CREATE TABLE debate (Date VARCHAR, Venue VARCHAR, Debate_ID VARCHAR); CREATE TABLE debate_people (Debate_ID VARCHAR, Affirmative VARCHAR)
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SELECT T3.Name, T2.Date, T2.Venue FROM debate_people AS T1 JOIN debate AS T2 ON T1.Debate_ID = T2.Debate_ID JOIN people AS T3 ON T1.Affirmative = T3.People_ID
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Mostre os nomes das pessoas, e as datas e locais dos debates que estão no lado negativo, ordenados em ordem alfabética ascendente de nome.
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CREATE TABLE people (Name VARCHAR, People_ID VARCHAR); CREATE TABLE debate (Date VARCHAR, Venue VARCHAR, Debate_ID VARCHAR); CREATE TABLE debate_people (Debate_ID VARCHAR, Negative VARCHAR)
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SELECT T3.Name, T2.Date, T2.Venue FROM debate_people AS T1 JOIN debate AS T2 ON T1.Debate_ID = T2.Debate_ID JOIN people AS T3 ON T1.Negative = T3.People_ID ORDER BY T3.Name
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Mostrar os nomes das pessoas que estão do lado positivo dos debates com número de audiência maior que 200.
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CREATE TABLE people (Name VARCHAR, People_ID VARCHAR); CREATE TABLE debate (Debate_ID VARCHAR, Num_of_Audience INTEGER); CREATE TABLE debate_people (Debate_ID VARCHAR, Affirmative VARCHAR)
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SELECT T3.Name FROM debate_people AS T1 JOIN debate AS T2 ON T1.Debate_ID = T2.Debate_ID JOIN people AS T3 ON T1.Affirmative = T3.People_ID WHERE T2.Num_of_Audience > 200
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Indique os nomes das pessoas e o número de vezes que estiveram do lado positivo dos debates.
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CREATE TABLE people (Name VARCHAR, People_ID VARCHAR); CREATE TABLE debate_people (Affirmative VARCHAR)
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SELECT T2.Name, COUNT(*) FROM debate_people AS T1 JOIN people AS T2 ON T1.Affirmative = T2.People_ID GROUP BY T2.Name
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Mostrar os nomes de pessoas que estiveram no lado negativo dos debates pelo menos duas vezes.
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CREATE TABLE debate_people (Negative VARCHAR); CREATE TABLE people (Name VARCHAR, People_ID VARCHAR)
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SELECT T2.Name FROM debate_people AS T1 JOIN people AS T2 ON T1.Negative = T2.People_ID GROUP BY T2.Name HAVING COUNT(*) >= 2
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Escreva os nomes de pessoas que não estiveram do lado positivo dos debates.
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CREATE TABLE debate_people (Name VARCHAR, People_id VARCHAR, Affirmative VARCHAR); CREATE TABLE people (Name VARCHAR, People_id VARCHAR, Affirmative VARCHAR)
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SELECT Name FROM people WHERE NOT People_id IN (SELECT Affirmative FROM debate_people)
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Escreva os nomes de todos os clientes em ordem alfabética.
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CREATE TABLE customers (customer_details VARCHAR)
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SELECT customer_details FROM customers ORDER BY customer_details
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Encontre todos os códigos do tipo de apólice associados ao cliente "Dayana Robel"
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CREATE TABLE customers (customer_id VARCHAR, customer_details VARCHAR); CREATE TABLE policies (customer_id VARCHAR)
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SELECT policy_type_code FROM policies AS t1 JOIN customers AS t2 ON t1.customer_id = t2.customer_id WHERE t2.customer_details = "Dayana Robel"
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Que tipo de política é mais frequentemente usado?
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CREATE TABLE policies (policy_type_code VARCHAR)
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SELECT policy_type_code FROM policies GROUP BY policy_type_code ORDER BY COUNT(*) DESC LIMIT 1
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Encontrar todos os tipos de políticas que são usados por mais de 2 clientes.
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CREATE TABLE policies (policy_type_code VARCHAR)
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SELECT policy_type_code FROM policies GROUP BY policy_type_code HAVING COUNT(*) > 2
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Encontre o valor total e médio pago nos cabeçalhos de reclamação.
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CREATE TABLE claim_headers (amount_piad INTEGER)
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SELECT SUM(amount_piad), AVG(amount_piad) FROM claim_headers
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Encontre o montante total reivindicado no documento mais recente.
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CREATE TABLE claim_headers (amount_claimed INTEGER, claim_header_id VARCHAR); CREATE TABLE claims_documents (claim_id VARCHAR, created_date VARCHAR); CREATE TABLE claims_documents (created_date VARCHAR)
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SELECT SUM(t1.amount_claimed) FROM claim_headers AS t1 JOIN claims_documents AS t2 ON t1.claim_header_id = t2.claim_id WHERE t2.created_date = (SELECT created_date FROM claims_documents ORDER BY created_date LIMIT 1)
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Qual é o nome do cliente que fez o maior montante de reclamação em uma única reclamação?
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CREATE TABLE customers (customer_details VARCHAR, customer_id VARCHAR); CREATE TABLE claim_headers (amount_claimed INTEGER); CREATE TABLE policies (policy_id VARCHAR, customer_id VARCHAR); CREATE TABLE claim_headers (policy_id VARCHAR, amount_claimed INTEGER)
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SELECT t3.customer_details FROM claim_headers AS t1 JOIN policies AS t2 ON t1.policy_id = t2.policy_id JOIN customers AS t3 ON t2.customer_id = t3.customer_id WHERE t1.amount_claimed = (SELECT MAX(amount_claimed) FROM claim_headers)
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Qual é o nome do cliente que efectuou o montante mínimo de pagamento num pedido?
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CREATE TABLE claim_headers (amount_piad INTEGER); CREATE TABLE customers (customer_details VARCHAR, customer_id VARCHAR); CREATE TABLE policies (policy_id VARCHAR, customer_id VARCHAR); CREATE TABLE claim_headers (policy_id VARCHAR, amount_piad INTEGER)
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SELECT t3.customer_details FROM claim_headers AS t1 JOIN policies AS t2 ON t1.policy_id = t2.policy_id JOIN customers AS t3 ON t2.customer_id = t3.customer_id WHERE t1.amount_piad = (SELECT MIN(amount_piad) FROM claim_headers)
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Encontre os nomes de clientes que não têm políticas associadas.
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CREATE TABLE customers (customer_details VARCHAR, customer_id VARCHAR); CREATE TABLE customers (customer_details VARCHAR); CREATE TABLE policies (customer_id VARCHAR)
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SELECT customer_details FROM customers EXCEPT SELECT t2.customer_details FROM policies AS t1 JOIN customers AS t2 ON t1.customer_id = t2.customer_id
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Quantas fases de tratamento de reclamações há no total?
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CREATE TABLE claims_processing_stages (Id VARCHAR)
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SELECT COUNT(*) FROM claims_processing_stages
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Qual é o nome da fase de processamento de reclamações em que a maioria das reclamações está?
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CREATE TABLE claims_processing (claim_stage_id VARCHAR); CREATE TABLE claims_processing_stages (claim_status_name VARCHAR, claim_stage_id VARCHAR)
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SELECT t2.claim_status_name FROM claims_processing AS t1 JOIN claims_processing_stages AS t2 ON t1.claim_stage_id = t2.claim_stage_id GROUP BY t1.claim_stage_id ORDER BY COUNT(*) DESC LIMIT 1
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Encontre os nomes de clientes cujo nome contém "Diana".
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CREATE TABLE customers (customer_details VARCHAR)
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SELECT customer_details FROM customers WHERE customer_details LIKE "%Diana%"
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Encontre os nomes dos clientes que têm uma apólice de substituição.
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CREATE TABLE policies (customer_id VARCHAR, policy_type_code VARCHAR); CREATE TABLE customers (customer_details VARCHAR, customer_id VARCHAR)
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SELECT DISTINCT t2.customer_details FROM policies AS t1 JOIN customers AS t2 ON t1.customer_id = t2.customer_id WHERE t1.policy_type_code = "Deputy"
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Encontre os nomes dos clientes que têm uma apólice de deputados ou de uniformes.
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CREATE TABLE policies (customer_id VARCHAR, policy_type_code VARCHAR); CREATE TABLE customers (customer_details VARCHAR, customer_id VARCHAR)
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SELECT DISTINCT t2.customer_details FROM policies AS t1 JOIN customers AS t2 ON t1.customer_id = t2.customer_id WHERE t1.policy_type_code = "Deputy" OR t1.policy_type_code = "Uniform"
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Encontre os nomes de todos os clientes e funcionários.
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CREATE TABLE staff (customer_details VARCHAR, staff_details VARCHAR); CREATE TABLE customers (customer_details VARCHAR, staff_details VARCHAR)
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SELECT customer_details FROM customers UNION SELECT staff_details FROM staff
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Determine o número de registos de cada tipo de política e o seu código de tipo.
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CREATE TABLE policies (policy_type_code VARCHAR)
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SELECT policy_type_code, COUNT(*) FROM policies GROUP BY policy_type_code
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Encontre o nome do cliente que esteve envolvido na maioria das apólices.
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CREATE TABLE customers (customer_details VARCHAR, customer_id VARCHAR); CREATE TABLE policies (customer_id VARCHAR)
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SELECT t2.customer_details FROM policies AS t1 JOIN customers AS t2 ON t1.customer_id = t2.customer_id GROUP BY t2.customer_details ORDER BY COUNT(*) DESC LIMIT 1
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Qual é a descrição do estado do sinistro "Abrido"?
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CREATE TABLE claims_processing_stages (claim_status_description VARCHAR, claim_status_name VARCHAR)
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SELECT claim_status_description FROM claims_processing_stages WHERE claim_status_name = "Open"
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Quantos códigos de resultado de reclamação diferentes há?
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CREATE TABLE claims_processing (claim_outcome_code VARCHAR)
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SELECT COUNT(DISTINCT claim_outcome_code) FROM claims_processing
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Que cliente está associado à última apólice?
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CREATE TABLE customers (customer_details VARCHAR, customer_id VARCHAR); CREATE TABLE policies (start_date INTEGER); CREATE TABLE policies (customer_id VARCHAR, start_date INTEGER)
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SELECT t2.customer_details FROM policies AS t1 JOIN customers AS t2 ON t1.customer_id = t2.customer_id WHERE t1.start_date = (SELECT MAX(start_date) FROM policies)
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Mostrar o id, a data de abertura da conta, o nome da conta e outros detalhes da conta para todas as contas.
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CREATE TABLE Accounts (account_id VARCHAR, date_account_opened VARCHAR, account_name VARCHAR, other_account_details VARCHAR)
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SELECT account_id, date_account_opened, account_name, other_account_details FROM Accounts
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Mostrar o número, o nome da conta e outros dados de todas as contas do cliente com o nome "Meaghan".
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CREATE TABLE Accounts (account_id VARCHAR, date_account_opened VARCHAR, account_name VARCHAR, other_account_details VARCHAR, customer_id VARCHAR); CREATE TABLE Customers (customer_id VARCHAR, customer_first_name VARCHAR)
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SELECT T1.account_id, T1.date_account_opened, T1.account_name, T1.other_account_details FROM Accounts AS T1 JOIN Customers AS T2 ON T1.customer_id = T2.customer_id WHERE T2.customer_first_name = 'Meaghan'
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Mostrar o nome da conta e outros detalhes de todas as contas do cliente com o nome de Meaghan e o sobrenome Keeling.
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CREATE TABLE Customers (customer_id VARCHAR, customer_first_name VARCHAR, customer_last_name VARCHAR); CREATE TABLE Accounts (account_name VARCHAR, other_account_details VARCHAR, customer_id VARCHAR)
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SELECT T1.account_name, T1.other_account_details FROM Accounts AS T1 JOIN Customers AS T2 ON T1.customer_id = T2.customer_id WHERE T2.customer_first_name = "Meaghan" AND T2.customer_last_name = "Keeling"
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Indicar o nome e o apelido do cliente com o nome da conta 900.
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CREATE TABLE Accounts (customer_id VARCHAR, account_name VARCHAR); CREATE TABLE Customers (customer_first_name VARCHAR, customer_last_name VARCHAR, customer_id VARCHAR)
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SELECT T2.customer_first_name, T2.customer_last_name FROM Accounts AS T1 JOIN Customers AS T2 ON T1.customer_id = T2.customer_id WHERE T1.account_name = "900"
|
Mostre os nomes únicos, sobrenomes e números de telefone de todos os clientes com qualquer conta.
|
CREATE TABLE Accounts (customer_id VARCHAR); CREATE TABLE Customers (customer_first_name VARCHAR, customer_last_name VARCHAR, phone_number VARCHAR, customer_id VARCHAR)
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SELECT DISTINCT T1.customer_first_name, T1.customer_last_name, T1.phone_number FROM Customers AS T1 JOIN Accounts AS T2 ON T1.customer_id = T2.customer_id
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Mostra os clientes que não têm uma conta.
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CREATE TABLE Customers (customer_id VARCHAR); CREATE TABLE Accounts (customer_id VARCHAR)
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SELECT customer_id FROM Customers EXCEPT SELECT customer_id FROM Accounts
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Quantas contas tem cada cliente?
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CREATE TABLE Accounts (customer_id VARCHAR)
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SELECT COUNT(*), customer_id FROM Accounts GROUP BY customer_id
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Qual é o número de clientes, nome e sobrenome com o maior número de contas.
|
CREATE TABLE Accounts (customer_id VARCHAR); CREATE TABLE Customers (customer_first_name VARCHAR, customer_last_name VARCHAR, customer_id VARCHAR)
|
SELECT T1.customer_id, T2.customer_first_name, T2.customer_last_name FROM Accounts AS T1 JOIN Customers AS T2 ON T1.customer_id = T2.customer_id GROUP BY T1.customer_id ORDER BY COUNT(*) DESC LIMIT 1
|
Mostrar a identificação, o nome e o apelido de todos os clientes e o número de contas.
|
CREATE TABLE Accounts (customer_id VARCHAR); CREATE TABLE Customers (customer_first_name VARCHAR, customer_last_name VARCHAR, customer_id VARCHAR)
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SELECT T1.customer_id, T2.customer_first_name, T2.customer_last_name, COUNT(*) FROM Accounts AS T1 JOIN Customers AS T2 ON T1.customer_id = T2.customer_id GROUP BY T1.customer_id
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Mostrar o nome e o número de identificação de todos os clientes com pelo menos 2 contas.
|
CREATE TABLE Customers (customer_first_name VARCHAR, customer_id VARCHAR); CREATE TABLE Accounts (customer_id VARCHAR)
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SELECT T2.customer_first_name, T1.customer_id FROM Accounts AS T1 JOIN Customers AS T2 ON T1.customer_id = T2.customer_id GROUP BY T1.customer_id HAVING COUNT(*) >= 2
|
Indicar o número de clientes para cada sexo.
|
CREATE TABLE Customers (gender VARCHAR)
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SELECT gender, COUNT(*) FROM Customers GROUP BY gender
|
Quantas transacções temos?
|
CREATE TABLE Financial_transactions (Id VARCHAR)
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SELECT COUNT(*) FROM Financial_transactions
|
Quantas transacções tem cada conta?
|
CREATE TABLE Financial_transactions (account_id VARCHAR)
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SELECT COUNT(*), account_id FROM Financial_transactions
|
Quantas transacções tem a conta com o nome 337?
|
CREATE TABLE Accounts (account_id VARCHAR, account_name VARCHAR); CREATE TABLE Financial_transactions (account_id VARCHAR)
|
SELECT COUNT(*) FROM Financial_transactions AS T1 JOIN Accounts AS T2 ON T1.account_id = T2.account_id WHERE T2.account_name = "337"
|
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