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Quais são os nomes dos países com a 3a maior população?
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CREATE TABLE country (Name VARCHAR, Population VARCHAR)
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SELECT Name FROM country ORDER BY Population DESC LIMIT 3
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Quais são os nomes das nações com a população mais baixa?
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CREATE TABLE country (Name VARCHAR, Population VARCHAR)
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SELECT Name FROM country ORDER BY Population LIMIT 3
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Quantos países há na Ásia?
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CREATE TABLE country (continent VARCHAR)
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SELECT COUNT(*) FROM country WHERE continent = "Asia"
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Quais são os nomes dos países que estão no continente europeu e têm uma população de 80000?
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CREATE TABLE country (Name VARCHAR, continent VARCHAR, Population VARCHAR)
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SELECT Name FROM country WHERE continent = "Europe" AND Population = "80000"
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Qual é a população total e a área média dos países do continente da América do Norte cuja área é maior que 3000 ?
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CREATE TABLE country (population INTEGER, surfacearea INTEGER, continent VARCHAR)
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SELECT SUM(population), AVG(surfacearea) FROM country WHERE continent = "north america" AND surfacearea > 3000
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Quais são as cidades com população entre 160000 e 900000?
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CREATE TABLE city (name VARCHAR, Population INTEGER)
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SELECT name FROM city WHERE Population BETWEEN 160000 AND 900000
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Devolva os nomes das cidades que têm população entre 160000 e 900000 .
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CREATE TABLE city (name VARCHAR, population INTEGER)
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SELECT name FROM city WHERE population BETWEEN 160000 AND 900000
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Qual a língua falada pelo maior número de países?
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CREATE TABLE countrylanguage (LANGUAGE VARCHAR)
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SELECT LANGUAGE FROM countrylanguage GROUP BY LANGUAGE ORDER BY COUNT(*) DESC LIMIT 1
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Qual é a língua falada pela maior percentagem de pessoas em cada país?
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CREATE TABLE countrylanguage (LANGUAGE VARCHAR, CountryCode VARCHAR, Percentage INTEGER)
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SELECT LANGUAGE, CountryCode, MAX(Percentage) FROM countrylanguage GROUP BY CountryCode
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Qual é o número total de países onde o espanhol é falado pela maior porcentagem de pessoas?
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CREATE TABLE countrylanguage (Percentage INTEGER, CountryCode VARCHAR, LANGUAGE VARCHAR)
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SELECT COUNT(*), MAX(Percentage) FROM countrylanguage WHERE LANGUAGE = "Spanish" GROUP BY CountryCode
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Quais são os códigos dos países onde o espanhol é falado pela maior porcentagem de pessoas?
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CREATE TABLE countrylanguage (CountryCode VARCHAR, Percentage INTEGER, LANGUAGE VARCHAR)
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SELECT CountryCode, MAX(Percentage) FROM countrylanguage WHERE LANGUAGE = "Spanish" GROUP BY CountryCode
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Quantos condutores há?
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CREATE TABLE conductor (Id VARCHAR)
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SELECT COUNT(*) FROM conductor
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Escreva os nomes dos condutores em ordem ascendente de idade.
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CREATE TABLE conductor (Name VARCHAR, Age VARCHAR)
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SELECT Name FROM conductor ORDER BY Age
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Quais são os nomes dos condutores cujas nacionalidades não são "EUA"?
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CREATE TABLE conductor (Name VARCHAR, Nationality VARCHAR)
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SELECT Name FROM conductor WHERE Nationality <> 'USA'
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Quais são as companhias discográficas de orquestras em ordem decrescente de anos em que foram fundadas?
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CREATE TABLE orchestra (Record_Company VARCHAR, Year_of_Founded VARCHAR)
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SELECT Record_Company FROM orchestra ORDER BY Year_of_Founded DESC
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Qual é a média de assistência a espetáculos?
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CREATE TABLE SHOW (Attendance INTEGER)
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SELECT AVG(Attendance) FROM SHOW
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Qual é a parcela máxima e mínima de performances cujo tipo não é "Live final"?
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CREATE TABLE performance (SHARE INTEGER, TYPE VARCHAR)
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SELECT MAX(SHARE), MIN(SHARE) FROM performance WHERE TYPE <> "Live final"
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Quantas nacionalidades diferentes têm os condutores?
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CREATE TABLE conductor (Nationality VARCHAR)
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SELECT COUNT(DISTINCT Nationality) FROM conductor
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Lista os nomes dos condutores em ordem decrescente de anos de trabalho.
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CREATE TABLE conductor (Name VARCHAR, Year_of_Work VARCHAR)
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SELECT Name FROM conductor ORDER BY Year_of_Work DESC
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Escreva o nome do condutor com mais anos de trabalho.
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CREATE TABLE conductor (Name VARCHAR, Year_of_Work VARCHAR)
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SELECT Name FROM conductor ORDER BY Year_of_Work DESC LIMIT 1
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Indique os nomes dos dirigentes e das orquestras que dirigiram.
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CREATE TABLE conductor (Name VARCHAR, Conductor_ID VARCHAR); CREATE TABLE orchestra (Orchestra VARCHAR, Conductor_ID VARCHAR)
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SELECT T1.Name, T2.Orchestra FROM conductor AS T1 JOIN orchestra AS T2 ON T1.Conductor_ID = T2.Conductor_ID
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Indique os nomes dos maestros que dirigiram mais de uma orquestra.
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CREATE TABLE orchestra (Conductor_ID VARCHAR); CREATE TABLE conductor (Name VARCHAR, Conductor_ID VARCHAR)
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SELECT T1.Name FROM conductor AS T1 JOIN orchestra AS T2 ON T1.Conductor_ID = T2.Conductor_ID GROUP BY T2.Conductor_ID HAVING COUNT(*) > 1
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Indicar o nome do maestro que dirigiu o maior número de orquestras.
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CREATE TABLE orchestra (Conductor_ID VARCHAR); CREATE TABLE conductor (Name VARCHAR, Conductor_ID VARCHAR)
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SELECT T1.Name FROM conductor AS T1 JOIN orchestra AS T2 ON T1.Conductor_ID = T2.Conductor_ID GROUP BY T2.Conductor_ID ORDER BY COUNT(*) DESC LIMIT 1
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Indicar o nome do maestro que tenha dirigido orquestras fundadas após 2008.
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CREATE TABLE orchestra (Conductor_ID VARCHAR); CREATE TABLE conductor (Name VARCHAR, Conductor_ID VARCHAR)
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SELECT T1.Name FROM conductor AS T1 JOIN orchestra AS T2 ON T1.Conductor_ID = T2.Conductor_ID WHERE Year_of_Founded > 2008
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Indique as diferentes gravadoras e o número correspondente de orquestras.
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CREATE TABLE orchestra (Record_Company VARCHAR)
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SELECT Record_Company, COUNT(*) FROM orchestra GROUP BY Record_Company
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Mostre os formatos de gravação das orquestras em ordem ascendente de contagem.
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CREATE TABLE orchestra (Major_Record_Format VARCHAR)
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SELECT Major_Record_Format FROM orchestra GROUP BY Major_Record_Format ORDER BY COUNT(*)
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Escreva a gravadora compartilhada pelo maior número de orquestras.
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CREATE TABLE orchestra (Record_Company VARCHAR)
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SELECT Record_Company FROM orchestra GROUP BY Record_Company ORDER BY COUNT(*) DESC LIMIT 1
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Escreva os nomes das orquestras que não têm apresentação.
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CREATE TABLE orchestra (Orchestra VARCHAR, Orchestra_ID VARCHAR); CREATE TABLE performance (Orchestra VARCHAR, Orchestra_ID VARCHAR)
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SELECT Orchestra FROM orchestra WHERE NOT Orchestra_ID IN (SELECT Orchestra_ID FROM performance)
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Mostrar as gravadoras partilhadas por orquestras fundadas antes de 2003 e depois de 2003.
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CREATE TABLE orchestra (Record_Company VARCHAR, Year_of_Founded INTEGER)
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SELECT Record_Company FROM orchestra WHERE Year_of_Founded < 2003 INTERSECT SELECT Record_Company FROM orchestra WHERE Year_of_Founded > 2003
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Encontre o número de orquestras cujo formato de gravação é "CD" ou "DVD".
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CREATE TABLE orchestra (Major_Record_Format VARCHAR)
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SELECT COUNT(*) FROM orchestra WHERE Major_Record_Format = "CD" OR Major_Record_Format = "DVD"
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Indicar os anos em que foram fundadas as orquestras que tiverem dado mais de uma apresentação.
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CREATE TABLE performance (Orchestra_ID VARCHAR); CREATE TABLE orchestra (Orchestra_ID VARCHAR)
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SELECT Year_of_Founded FROM orchestra AS T1 JOIN performance AS T2 ON T1.Orchestra_ID = T2.Orchestra_ID GROUP BY T2.Orchestra_ID HAVING COUNT(*) > 1
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Quantos estudantes de secundário há?
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CREATE TABLE Highschooler (Id VARCHAR)
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SELECT COUNT(*) FROM Highschooler
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Mostre os nomes e notas de cada estudante do ensino médio.
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CREATE TABLE Highschooler (name VARCHAR, grade VARCHAR)
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SELECT name, grade FROM Highschooler
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Mostra as notas dos alunos do liceu.
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CREATE TABLE Highschooler (grade VARCHAR)
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SELECT grade FROM Highschooler
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Em que ano está o Kyle?
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CREATE TABLE Highschooler (grade VARCHAR, name VARCHAR)
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SELECT grade FROM Highschooler WHERE name = "Kyle"
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Mostra os nomes de todos os alunos do secundário do 10o ano.
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CREATE TABLE Highschooler (name VARCHAR, grade VARCHAR)
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SELECT name FROM Highschooler WHERE grade = 10
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Mostra a identificação do estudante do liceu chamado Kyle.
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CREATE TABLE Highschooler (ID VARCHAR, name VARCHAR)
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SELECT ID FROM Highschooler WHERE name = "Kyle"
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Quantos alunos do liceu estão no 9o ou 10o ano?
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CREATE TABLE Highschooler (grade VARCHAR)
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SELECT COUNT(*) FROM Highschooler WHERE grade = 9 OR grade = 10
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Mostre o número de estudantes do ensino médio para cada ano.
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CREATE TABLE Highschooler (grade VARCHAR)
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SELECT grade, COUNT(*) FROM Highschooler GROUP BY grade
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Qual a classe que tem mais alunos do secundário?
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CREATE TABLE Highschooler (grade VARCHAR)
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SELECT grade FROM Highschooler GROUP BY grade ORDER BY COUNT(*) DESC LIMIT 1
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Mostre-me todas as notas que têm pelo menos 4 alunos.
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CREATE TABLE Highschooler (grade VARCHAR)
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SELECT grade FROM Highschooler GROUP BY grade HAVING COUNT(*) >= 4
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Mostre os documentos dos alunos e o número de amigos correspondentes a cada um.
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CREATE TABLE Friend (student_id VARCHAR)
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SELECT student_id, COUNT(*) FROM Friend GROUP BY student_id
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Mostre os nomes dos estudantes do ensino médio e o número correspondente de amigos.
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CREATE TABLE Highschooler (name VARCHAR, id VARCHAR); CREATE TABLE Friend (student_id VARCHAR)
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SELECT T2.name, COUNT(*) FROM Friend AS T1 JOIN Highschooler AS T2 ON T1.student_id = T2.id GROUP BY T1.student_id
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Qual é o nome do estudante do ensino médio que tem o maior número de amigos?
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CREATE TABLE Highschooler (name VARCHAR, id VARCHAR); CREATE TABLE Friend (student_id VARCHAR)
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SELECT T2.name FROM Friend AS T1 JOIN Highschooler AS T2 ON T1.student_id = T2.id GROUP BY T1.student_id ORDER BY COUNT(*) DESC LIMIT 1
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Mostre os nomes de estudantes do ensino secundário que têm pelo menos 3 amigos.
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CREATE TABLE Highschooler (name VARCHAR, id VARCHAR); CREATE TABLE Friend (student_id VARCHAR)
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SELECT T2.name FROM Friend AS T1 JOIN Highschooler AS T2 ON T1.student_id = T2.id GROUP BY T1.student_id HAVING COUNT(*) >= 3
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Mostra os nomes de todos os amigos do Kyle.
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CREATE TABLE Highschooler (name VARCHAR, id VARCHAR); CREATE TABLE Friend (student_id VARCHAR, friend_id VARCHAR); CREATE TABLE Highschooler (id VARCHAR, name VARCHAR)
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SELECT T3.name FROM Friend AS T1 JOIN Highschooler AS T2 ON T1.student_id = T2.id JOIN Highschooler AS T3 ON T1.friend_id = T3.id WHERE T2.name = "Kyle"
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Quantos amigos tem o estudante do liceu Kyle?
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CREATE TABLE Friend (student_id VARCHAR); CREATE TABLE Highschooler (id VARCHAR, name VARCHAR)
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SELECT COUNT(*) FROM Friend AS T1 JOIN Highschooler AS T2 ON T1.student_id = T2.id WHERE T2.name = "Kyle"
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Mostre identificação de todos os estudantes que não têm amigos.
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CREATE TABLE Highschooler (id VARCHAR, student_id VARCHAR); CREATE TABLE Friend (id VARCHAR, student_id VARCHAR)
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SELECT id FROM Highschooler EXCEPT SELECT student_id FROM Friend
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Mostre os nomes de todos os estudantes do ensino médio que não têm amigos.
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CREATE TABLE Highschooler (name VARCHAR, id VARCHAR); CREATE TABLE Highschooler (name VARCHAR); CREATE TABLE Friend (student_id VARCHAR)
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SELECT name FROM Highschooler EXCEPT SELECT T2.name FROM Friend AS T1 JOIN Highschooler AS T2 ON T1.student_id = T2.id
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Mostra os documentos dos alunos do secundário que têm amigos e que também são gostados por outra pessoa.
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CREATE TABLE Likes (student_id VARCHAR, liked_id VARCHAR); CREATE TABLE Friend (student_id VARCHAR, liked_id VARCHAR)
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SELECT student_id FROM Friend INTERSECT SELECT liked_id FROM Likes
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Mostrar o nome de todos os alunos que têm alguns amigos e também são gostados por alguém.
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CREATE TABLE Highschooler (name VARCHAR, id VARCHAR); CREATE TABLE Likes (student_id VARCHAR, liked_id VARCHAR); CREATE TABLE Friend (student_id VARCHAR, liked_id VARCHAR)
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SELECT T2.name FROM Friend AS T1 JOIN Highschooler AS T2 ON T1.student_id = T2.id INTERSECT SELECT T2.name FROM Likes AS T1 JOIN Highschooler AS T2 ON T1.liked_id = T2.id
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Contar o número de curtidas para cada identificação de estudante.
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CREATE TABLE Likes (student_id VARCHAR)
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SELECT student_id, COUNT(*) FROM Likes GROUP BY student_id
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Mostrar os nomes dos estudantes do ensino secundário que têm curtidas e o número de curtidas para cada um.
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CREATE TABLE Likes (student_id VARCHAR); CREATE TABLE Highschooler (name VARCHAR, id VARCHAR)
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SELECT T2.name, COUNT(*) FROM Likes AS T1 JOIN Highschooler AS T2 ON T1.student_id = T2.id GROUP BY T1.student_id
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Qual é o nome do estudante do liceu que tem mais curtidas?
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CREATE TABLE Likes (student_id VARCHAR); CREATE TABLE Highschooler (name VARCHAR, id VARCHAR)
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SELECT T2.name FROM Likes AS T1 JOIN Highschooler AS T2 ON T1.student_id = T2.id GROUP BY T1.student_id ORDER BY COUNT(*) DESC LIMIT 1
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Mostre os nomes dos alunos que têm pelo menos 2 curtidas.
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CREATE TABLE Likes (student_id VARCHAR); CREATE TABLE Highschooler (name VARCHAR, id VARCHAR)
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SELECT T2.name FROM Likes AS T1 JOIN Highschooler AS T2 ON T1.student_id = T2.id GROUP BY T1.student_id HAVING COUNT(*) >= 2
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Mostrar os nomes dos alunos que têm uma nota superior a 5 e têm pelo menos 2 amigos.
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CREATE TABLE Friend (student_id VARCHAR); CREATE TABLE Highschooler (name VARCHAR, id VARCHAR, grade INTEGER)
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SELECT T2.name FROM Friend AS T1 JOIN Highschooler AS T2 ON T1.student_id = T2.id WHERE T2.grade > 5 GROUP BY T1.student_id HAVING COUNT(*) >= 2
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Quantos "likes" tem o Kyle?
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CREATE TABLE Likes (student_id VARCHAR); CREATE TABLE Highschooler (id VARCHAR, name VARCHAR)
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SELECT COUNT(*) FROM Likes AS T1 JOIN Highschooler AS T2 ON T1.student_id = T2.id WHERE T2.name = "Kyle"
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Encontre a nota média de todos os alunos que têm amigos.
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CREATE TABLE Highschooler (id VARCHAR); CREATE TABLE Friend (student_id VARCHAR); CREATE TABLE Highschooler (grade INTEGER, id VARCHAR)
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SELECT AVG(grade) FROM Highschooler WHERE id IN (SELECT T1.student_id FROM Friend AS T1 JOIN Highschooler AS T2 ON T1.student_id = T2.id)
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Encontre a nota mínima de estudantes que não têm amigos.
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CREATE TABLE Highschooler (id VARCHAR); CREATE TABLE Friend (student_id VARCHAR); CREATE TABLE Highschooler (grade INTEGER, id VARCHAR)
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SELECT MIN(grade) FROM Highschooler WHERE NOT id IN (SELECT T1.student_id FROM Friend AS T1 JOIN Highschooler AS T2 ON T1.student_id = T2.id)
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Que estados têm proprietários e profissionais que vivem lá?
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CREATE TABLE Owners (state VARCHAR); CREATE TABLE Professionals (state VARCHAR)
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SELECT state FROM Owners INTERSECT SELECT state FROM Professionals
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Qual é a idade média dos cães que foram tratados?
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CREATE TABLE Dogs (age INTEGER, dog_id VARCHAR); CREATE TABLE Treatments (age INTEGER, dog_id VARCHAR)
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SELECT AVG(age) FROM Dogs WHERE dog_id IN (SELECT dog_id FROM Treatments)
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Que profissionais vivem no estado de Indiana ou foram tratados em mais de 2 tratamentos?
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CREATE TABLE Treatments (professional_id VARCHAR); CREATE TABLE Professionals (professional_id VARCHAR, last_name VARCHAR, cell_number VARCHAR); CREATE TABLE Professionals (professional_id VARCHAR, last_name VARCHAR, cell_number VARCHAR, state VARCHAR)
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SELECT professional_id, last_name, cell_number FROM Professionals WHERE state = 'Indiana' UNION SELECT T1.professional_id, T1.last_name, T1.cell_number FROM Professionals AS T1 JOIN Treatments AS T2 ON T1.professional_id = T2.professional_id GROUP BY T1.professional_id HAVING COUNT(*) > 2
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Que cães não custaram ao seu dono mais de 1000 dólares por tratamento ?
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CREATE TABLE dogs (name VARCHAR, dog_id VARCHAR, cost_of_treatment INTEGER); CREATE TABLE treatments (name VARCHAR, dog_id VARCHAR, cost_of_treatment INTEGER)
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SELECT name FROM dogs WHERE NOT dog_id IN (SELECT dog_id FROM treatments GROUP BY dog_id HAVING SUM(cost_of_treatment) > 1000)
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Quais os nomes de primeiro nome que são usados para profissionais ou proprietários, mas não são usados como nomes de cães?
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CREATE TABLE Owners (first_name VARCHAR, name VARCHAR); CREATE TABLE Dogs (first_name VARCHAR, name VARCHAR); CREATE TABLE Professionals (first_name VARCHAR, name VARCHAR)
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SELECT first_name FROM Professionals UNION SELECT first_name FROM Owners EXCEPT SELECT name FROM Dogs
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Qual profissional não realizou nenhum tratamento em cães?
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CREATE TABLE Professionals (professional_id VARCHAR, role_code VARCHAR, email_address VARCHAR); CREATE TABLE Treatments (professional_id VARCHAR)
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SELECT professional_id, role_code, email_address FROM Professionals EXCEPT SELECT T1.professional_id, T1.role_code, T1.email_address FROM Professionals AS T1 JOIN Treatments AS T2 ON T1.professional_id = T2.professional_id
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Qual é o dono que possui mais cães?
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CREATE TABLE Owners (first_name VARCHAR, last_name VARCHAR, owner_id VARCHAR); CREATE TABLE Dogs (owner_id VARCHAR)
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SELECT T1.owner_id, T2.first_name, T2.last_name FROM Dogs AS T1 JOIN Owners AS T2 ON T1.owner_id = T2.owner_id GROUP BY T1.owner_id ORDER BY COUNT(*) DESC LIMIT 1
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Que profissionais já fizeram pelo menos dois tratamentos?
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CREATE TABLE Professionals (professional_id VARCHAR, role_code VARCHAR, first_name VARCHAR); CREATE TABLE Treatments (professional_id VARCHAR)
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SELECT T1.professional_id, T1.role_code, T1.first_name FROM Professionals AS T1 JOIN Treatments AS T2 ON T1.professional_id = T2.professional_id GROUP BY T1.professional_id HAVING COUNT(*) >= 2
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Qual é o nome da raça com mais cães?
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CREATE TABLE Dogs (breed_code VARCHAR); CREATE TABLE Breeds (breed_name VARCHAR, breed_code VARCHAR)
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SELECT T1.breed_name FROM Breeds AS T1 JOIN Dogs AS T2 ON T1.breed_code = T2.breed_code GROUP BY T1.breed_name ORDER BY COUNT(*) DESC LIMIT 1
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Qual é o dono que mais pagou pelos tratamentos dos cães?
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CREATE TABLE Owners (owner_id VARCHAR, last_name VARCHAR); CREATE TABLE Dogs (owner_id VARCHAR, dog_id VARCHAR); CREATE TABLE Treatments (dog_id VARCHAR)
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SELECT T1.owner_id, T1.last_name FROM Owners AS T1 JOIN Dogs AS T2 ON T1.owner_id = T2.owner_id JOIN Treatments AS T3 ON T2.dog_id = T3.dog_id GROUP BY T1.owner_id ORDER BY COUNT(*) DESC LIMIT 1
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Qual é a descrição do tipo de tratamento que custa menos dinheiro no total?
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CREATE TABLE Treatments (treatment_type_code VARCHAR); CREATE TABLE Treatment_types (treatment_type_description VARCHAR, treatment_type_code VARCHAR)
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SELECT T1.treatment_type_description FROM Treatment_types AS T1 JOIN Treatments AS T2 ON T1.treatment_type_code = T2.treatment_type_code GROUP BY T1.treatment_type_code ORDER BY SUM(cost_of_treatment) LIMIT 1
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Que dono pagou a maior quantia total pelos cães? Mostre o número e o código postal do dono.
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CREATE TABLE Treatments (dog_id VARCHAR, cost_of_treatment INTEGER); CREATE TABLE Owners (owner_id VARCHAR, zip_code VARCHAR); CREATE TABLE Dogs (owner_id VARCHAR, dog_id VARCHAR)
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SELECT T1.owner_id, T1.zip_code FROM Owners AS T1 JOIN Dogs AS T2 ON T1.owner_id = T2.owner_id JOIN Treatments AS T3 ON T2.dog_id = T3.dog_id GROUP BY T1.owner_id ORDER BY SUM(T3.cost_of_treatment) DESC LIMIT 1
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Que profissionais fizeram pelo menos dois tipos de tratamentos?
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CREATE TABLE Professionals (professional_id VARCHAR, cell_number VARCHAR); CREATE TABLE Treatments (professional_id VARCHAR)
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SELECT T1.professional_id, T1.cell_number FROM Professionals AS T1 JOIN Treatments AS T2 ON T1.professional_id = T2.professional_id GROUP BY T1.professional_id HAVING COUNT(*) >= 2
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Qual é o nome e o apelido dos profissionais que fizeram tratamentos com custos abaixo da média?
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CREATE TABLE Treatments (cost_of_treatment INTEGER); CREATE TABLE Professionals (first_name VARCHAR, last_name VARCHAR); CREATE TABLE Treatments (Id VARCHAR)
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SELECT DISTINCT T1.first_name, T1.last_name FROM Professionals AS T1 JOIN Treatments AS T2 WHERE cost_of_treatment < (SELECT AVG(cost_of_treatment) FROM Treatments)
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Indique a data de cada tratamento, juntamente com o nome do profissional que o realizou.
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CREATE TABLE Treatments (date_of_treatment VARCHAR, professional_id VARCHAR); CREATE TABLE Professionals (first_name VARCHAR, professional_id VARCHAR)
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SELECT T1.date_of_treatment, T2.first_name FROM Treatments AS T1 JOIN Professionals AS T2 ON T1.professional_id = T2.professional_id
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Indicar o custo de cada tratamento e a descrição do tipo de tratamento correspondente.
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CREATE TABLE Treatments (cost_of_treatment VARCHAR, treatment_type_code VARCHAR); CREATE TABLE treatment_types (treatment_type_description VARCHAR, treatment_type_code VARCHAR)
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SELECT T1.cost_of_treatment, T2.treatment_type_description FROM Treatments AS T1 JOIN treatment_types AS T2 ON T1.treatment_type_code = T2.treatment_type_code
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Escreva o nome de cada dono, o apelido e o tamanho do seu para o cão.
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CREATE TABLE Owners (first_name VARCHAR, last_name VARCHAR, owner_id VARCHAR); CREATE TABLE Dogs (size_code VARCHAR, owner_id VARCHAR)
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SELECT T1.first_name, T1.last_name, T2.size_code FROM Owners AS T1 JOIN Dogs AS T2 ON T1.owner_id = T2.owner_id
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Lista de pares de nome do dono e do nome do cão.
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CREATE TABLE Dogs (name VARCHAR, owner_id VARCHAR); CREATE TABLE Owners (first_name VARCHAR, owner_id VARCHAR)
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SELECT T1.first_name, T2.name FROM Owners AS T1 JOIN Dogs AS T2 ON T1.owner_id = T2.owner_id
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Escreva os nomes dos cães da raça mais rara e as datas de tratamento deles.
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CREATE TABLE Dogs (name VARCHAR, dog_id VARCHAR, breed_code VARCHAR); CREATE TABLE Treatments (date_of_treatment VARCHAR, dog_id VARCHAR); CREATE TABLE Dogs (breed_code VARCHAR)
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SELECT T1.name, T2.date_of_treatment FROM Dogs AS T1 JOIN Treatments AS T2 ON T1.dog_id = T2.dog_id WHERE T1.breed_code = (SELECT breed_code FROM Dogs GROUP BY breed_code ORDER BY COUNT(*) LIMIT 1)
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Que cães são de alguém que mora na Virgínia?
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CREATE TABLE Dogs (name VARCHAR, owner_id VARCHAR); CREATE TABLE Owners (first_name VARCHAR, owner_id VARCHAR, state VARCHAR)
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SELECT T1.first_name, T2.name FROM Owners AS T1 JOIN Dogs AS T2 ON T1.owner_id = T2.owner_id WHERE T1.state = 'Virginia'
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Qual é a data de chegada e a data de partida dos cães que tiverem sido tratados?
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CREATE TABLE Dogs (date_arrived VARCHAR, date_departed VARCHAR, dog_id VARCHAR); CREATE TABLE Treatments (dog_id VARCHAR)
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SELECT DISTINCT T1.date_arrived, T1.date_departed FROM Dogs AS T1 JOIN Treatments AS T2 ON T1.dog_id = T2.dog_id
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Escreva o sobrenome do dono do cão mais novo.
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CREATE TABLE Owners (last_name VARCHAR, owner_id VARCHAR); CREATE TABLE Dogs (owner_id VARCHAR, age INTEGER); CREATE TABLE Dogs (age INTEGER)
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SELECT T1.last_name FROM Owners AS T1 JOIN Dogs AS T2 ON T1.owner_id = T2.owner_id WHERE T2.age = (SELECT MAX(age) FROM Dogs)
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Escreva os e-mails dos profissionais que vivem no estado do Havai ou no estado de Wisconsin.
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CREATE TABLE Professionals (email_address VARCHAR, state VARCHAR)
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SELECT email_address FROM Professionals WHERE state = 'Hawaii' OR state = 'Wisconsin'
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Qual é a data de chegada e a data de partida de todos os cães?
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CREATE TABLE Dogs (date_arrived VARCHAR, date_departed VARCHAR)
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SELECT date_arrived, date_departed FROM Dogs
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Quantos cães foram tratados?
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CREATE TABLE Treatments (dog_id VARCHAR)
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SELECT COUNT(DISTINCT dog_id) FROM Treatments
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Quantos profissionais já realizaram algum tratamento a cães?
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CREATE TABLE Treatments (professional_id VARCHAR)
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SELECT COUNT(DISTINCT professional_id) FROM Treatments
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Que profissionais vivem numa cidade que contenha a substring "Oeste"?
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CREATE TABLE professionals (role_code VARCHAR, street VARCHAR, city VARCHAR, state VARCHAR)
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SELECT role_code, street, city, state FROM professionals WHERE city LIKE '%West%'
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Que proprietários vivem no estado cujo nome contém a substring "North"?
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CREATE TABLE Owners (first_name VARCHAR, last_name VARCHAR, email_address VARCHAR, state VARCHAR)
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SELECT first_name, last_name, email_address FROM Owners WHERE state LIKE '%North%'
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Quantos cães têm uma idade abaixo da média?
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CREATE TABLE Dogs (age INTEGER)
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SELECT COUNT(*) FROM Dogs WHERE age < (SELECT AVG(age) FROM Dogs)
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Quanto custa o tratamento mais recente?
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CREATE TABLE Treatments (cost_of_treatment VARCHAR, date_of_treatment VARCHAR)
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SELECT cost_of_treatment FROM Treatments ORDER BY date_of_treatment DESC LIMIT 1
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Quantos cães não foram tratados?
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CREATE TABLE Dogs (dog_id VARCHAR); CREATE TABLE Treatments (dog_id VARCHAR)
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SELECT COUNT(*) FROM Dogs WHERE NOT dog_id IN (SELECT dog_id FROM Treatments)
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Diga-me quantos cães não receberam tratamento .
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CREATE TABLE treatments (dog_id VARCHAR); CREATE TABLE dogs (dog_id VARCHAR)
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SELECT COUNT(*) FROM dogs WHERE NOT dog_id IN (SELECT dog_id FROM treatments)
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Quantos donos não têm cães temporariamente?
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CREATE TABLE Dogs (owner_id VARCHAR); CREATE TABLE Owners (owner_id VARCHAR)
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SELECT COUNT(*) FROM Owners WHERE NOT owner_id IN (SELECT owner_id FROM Dogs)
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Quantos profissionais não realizaram nenhum tratamento em cães?
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CREATE TABLE Professionals (professional_id VARCHAR); CREATE TABLE Treatments (professional_id VARCHAR)
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SELECT COUNT(*) FROM Professionals WHERE NOT professional_id IN (SELECT professional_id FROM Treatments)
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Escreva o nome, idade e peso dos cães que foram abandonados? 1 significa sim e 0 significa não.
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CREATE TABLE Dogs (name VARCHAR, age VARCHAR, weight VARCHAR, abandoned_yn VARCHAR)
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SELECT name, age, weight FROM Dogs WHERE abandoned_yn = 1
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Qual é a idade média de todos os cães?
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CREATE TABLE Dogs (age INTEGER)
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SELECT AVG(age) FROM Dogs
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Qual é a idade do cão mais velho?
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CREATE TABLE Dogs (age INTEGER)
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SELECT MAX(age) FROM Dogs
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Quanto custa cada tipo de cobrança?
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CREATE TABLE Charges (charge_type VARCHAR, charge_amount VARCHAR)
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SELECT charge_type, charge_amount FROM Charges
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Quanto custa o tipo de carregamento mais caro?
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CREATE TABLE Charges (charge_amount INTEGER)
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SELECT MAX(charge_amount) FROM Charges
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Escreva o e-mail, o telemóvel e o telefone de casa de todos os profissionais.
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CREATE TABLE professionals (email_address VARCHAR, cell_number VARCHAR, home_phone VARCHAR)
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SELECT email_address, cell_number, home_phone FROM professionals
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Quais são todas as combinações possíveis de tipo de raça e tamanho?
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CREATE TABLE dogs (breed_code VARCHAR, size_code VARCHAR)
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SELECT DISTINCT breed_code, size_code FROM dogs
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