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sql-create-context-pt

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Determine as datas de envio dos documentos com o montante de subvenção superior a 5000 foram concedidos por tipo de organização descrito
CREATE TABLE organisation_Types (organisation_type VARCHAR, organisation_type_description VARCHAR); CREATE TABLE Grants (grant_id VARCHAR, organisation_id VARCHAR, grant_amount VARCHAR); CREATE TABLE Organisations (organisation_id VARCHAR, organisation_type VARCHAR); CREATE TABLE documents (sent_date VARCHAR, grant_id VARCHAR)
SELECT T1.sent_date FROM documents AS T1 JOIN Grants AS T2 ON T1.grant_id = T2.grant_id JOIN Organisations AS T3 ON T2.organisation_id = T3.organisation_id JOIN organisation_Types AS T4 ON T3.organisation_type = T4.organisation_type WHERE T2.grant_amount > 5000 AND T4.organisation_type_description = 'Research'
Qual é a data de recepção das respostas para os documentos descritos como "regulares" ou concedidos com mais de 100?
CREATE TABLE Documents (response_received_date VARCHAR, document_type_code VARCHAR, grant_id VARCHAR); CREATE TABLE Document_Types (document_type_code VARCHAR, document_description VARCHAR); CREATE TABLE Grants (grant_id VARCHAR, grant_amount VARCHAR)
SELECT T1.response_received_date FROM Documents AS T1 JOIN Document_Types AS T2 ON T1.document_type_code = T2.document_type_code JOIN Grants AS T3 ON T1.grant_id = T3.grant_id WHERE T2.document_description = 'Regular' OR T3.grant_amount > 100
A lista dos projectos que não contrataram pessoal para um papel de investigador.
CREATE TABLE Projects (project_details VARCHAR, project_id VARCHAR, role_code VARCHAR); CREATE TABLE Project_Staff (project_details VARCHAR, project_id VARCHAR, role_code VARCHAR)
SELECT project_details FROM Projects WHERE NOT project_id IN (SELECT project_id FROM Project_Staff WHERE role_code = 'researcher')
Quais são os detalhes das tarefas, o ID da tarefa e o ID do projeto para os projetos que são descritos como "omnis" ou que têm mais de 2 resultados?
CREATE TABLE Projects (project_id VARCHAR, project_details VARCHAR); CREATE TABLE Project_outcomes (project_id VARCHAR); CREATE TABLE Tasks (task_details VARCHAR, task_id VARCHAR, project_id VARCHAR)
SELECT T1.task_details, T1.task_id, T2.project_id FROM Tasks AS T1 JOIN Projects AS T2 ON T1.project_id = T2.project_id WHERE T2.project_details = 'omnis' UNION SELECT T1.task_details, T1.task_id, T2.project_id FROM Tasks AS T1 JOIN Projects AS T2 ON T1.project_id = T2.project_id JOIN Project_outcomes AS T3 ON T2.project_id = T3.project_id GROUP BY T2.project_id HAVING COUNT(*) > 2
Quando é que todo o pessoal do papel de pesquisador começa a trabalhar e quando é que eles param de trabalhar?
CREATE TABLE Project_Staff (date_from VARCHAR, date_to VARCHAR, role_code VARCHAR)
SELECT date_from, date_to FROM Project_Staff WHERE role_code = 'researcher'
Quantos tipos de papéis há para o pessoal?
CREATE TABLE Project_Staff (role_code VARCHAR)
SELECT COUNT(DISTINCT role_code) FROM Project_Staff
Qual é o montante total das subvenções concedidas por cada organização?
CREATE TABLE Grants (organisation_id VARCHAR, grant_amount INTEGER)
SELECT SUM(grant_amount), organisation_id FROM Grants GROUP BY organisation_id
A lista dos projectos com o resultado da investigação descrito com a substring "Publicado".
CREATE TABLE Research_outcomes (outcome_code VARCHAR, outcome_description VARCHAR); CREATE TABLE Project_outcomes (project_id VARCHAR, outcome_code VARCHAR); CREATE TABLE Projects (project_details VARCHAR, project_id VARCHAR)
SELECT T1.project_details FROM Projects AS T1 JOIN Project_outcomes AS T2 ON T1.project_id = T2.project_id JOIN Research_outcomes AS T3 ON T2.outcome_code = T3.outcome_code WHERE T3.outcome_description LIKE '%Published%'
Quantos funcionários tem cada projeto?
CREATE TABLE Project_Staff (project_id VARCHAR); CREATE TABLE Projects (project_id VARCHAR)
SELECT T1.project_id, COUNT(*) FROM Project_Staff AS T1 JOIN Projects AS T2 ON T1.project_id = T2.project_id GROUP BY T1.project_id ORDER BY COUNT(*)
Qual é a descrição completa do papel do investigador?
CREATE TABLE Staff_Roles (role_description VARCHAR, role_code VARCHAR)
SELECT role_description FROM Staff_Roles WHERE role_code = 'researcher'
Quando é que o primeiro funcionário para os projetos começou a trabalhar?
CREATE TABLE Project_Staff (date_from VARCHAR)
SELECT date_from FROM Project_Staff ORDER BY date_from LIMIT 1
Qual o projeto que obteve mais resultados?
CREATE TABLE Project_outcomes (project_id VARCHAR); CREATE TABLE Projects (project_details VARCHAR, project_id VARCHAR)
SELECT T1.project_details, T1.project_id FROM Projects AS T1 JOIN Project_outcomes AS T2 ON T1.project_id = T2.project_id GROUP BY T1.project_id ORDER BY COUNT(*) DESC LIMIT 1
Que projetos não têm resultado?
CREATE TABLE Project_outcomes (project_details VARCHAR, project_id VARCHAR); CREATE TABLE Projects (project_details VARCHAR, project_id VARCHAR)
SELECT project_details FROM Projects WHERE NOT project_id IN (SELECT project_id FROM Project_outcomes)
Qual a organização que contratou mais pessoal de investigação?
CREATE TABLE Organisations (organisation_id VARCHAR, organisation_type VARCHAR, organisation_details VARCHAR); CREATE TABLE Research_Staff (employer_organisation_id VARCHAR)
SELECT T1.organisation_id, T1.organisation_type, T1.organisation_details FROM Organisations AS T1 JOIN Research_Staff AS T2 ON T1.organisation_id = T2.employer_organisation_id GROUP BY T1.organisation_id ORDER BY COUNT(*) DESC LIMIT 1
Indicar a descrição das funções e a identificação do pessoal do projeto envolvido na maioria dos resultados do projeto?
CREATE TABLE Project_outcomes (project_id VARCHAR); CREATE TABLE Project_Staff (staff_id VARCHAR, role_code VARCHAR, project_id VARCHAR); CREATE TABLE Staff_Roles (role_description VARCHAR, role_code VARCHAR)
SELECT T1.role_description, T2.staff_id FROM Staff_Roles AS T1 JOIN Project_Staff AS T2 ON T1.role_code = T2.role_code JOIN Project_outcomes AS T3 ON T2.project_id = T3.project_id GROUP BY T2.staff_id ORDER BY COUNT(*) DESC LIMIT 1
Qual o tipo de documento descrito com o prefixo "Initial"?
CREATE TABLE Document_Types (document_type_code VARCHAR, document_description VARCHAR)
SELECT document_type_code FROM Document_Types WHERE document_description LIKE 'Initial%'
Para as subvenções com os dois documentos descritos como "Regular" e "Initial Application", indicar a data de início.
CREATE TABLE Grants (grant_start_date VARCHAR, grant_id VARCHAR); CREATE TABLE Document_Types (document_type_code VARCHAR, document_description VARCHAR); CREATE TABLE Documents (grant_id VARCHAR, document_type_code VARCHAR)
SELECT T1.grant_start_date FROM Grants AS T1 JOIN Documents AS T2 ON T1.grant_id = T2.grant_id JOIN Document_Types AS T3 ON T2.document_type_code = T3.document_type_code WHERE T3.document_description = 'Regular' INTERSECT SELECT T1.grant_start_date FROM Grants AS T1 JOIN Documents AS T2 ON T1.grant_id = T2.grant_id JOIN Document_Types AS T3 ON T2.document_type_code = T3.document_type_code WHERE T3.document_description = 'Initial Application'
Quantos documentos pode ter uma subvenção no máximo?
CREATE TABLE Documents (grant_id VARCHAR)
SELECT grant_id, COUNT(*) FROM Documents GROUP BY grant_id ORDER BY COUNT(*) DESC LIMIT 1
Encontre a descrição do tipo de organização da organização, detalhada como "quo".
CREATE TABLE Organisations (organisation_type VARCHAR, organisation_details VARCHAR); CREATE TABLE organisation_Types (organisation_type_description VARCHAR, organisation_type VARCHAR)
SELECT T1.organisation_type_description FROM organisation_Types AS T1 JOIN Organisations AS T2 ON T1.organisation_type = T2.organisation_type WHERE T2.organisation_details = 'quo'
Qual é o número de organizações descritas como "Proponente"?
CREATE TABLE organisation_Types (organisation_type VARCHAR, organisation_type_description VARCHAR); CREATE TABLE Organisations (organisation_type VARCHAR)
SELECT organisation_details FROM Organisations AS T1 JOIN organisation_Types AS T2 ON T1.organisation_type = T2.organisation_type WHERE T2.organisation_type_description = 'Sponsor' ORDER BY organisation_details
Quantos resultados de patentes são gerados a partir de todos os projetos?
CREATE TABLE Project_outcomes (outcome_code VARCHAR)
SELECT COUNT(*) FROM Project_outcomes WHERE outcome_code = 'Patent'
Quantos funcionários do projeto trabalharam como líderes ou começaram a trabalhar antes de '1989-04-24 23:51:54'?
CREATE TABLE Project_Staff (role_code VARCHAR, date_from VARCHAR)
SELECT COUNT(*) FROM Project_Staff WHERE role_code = 'leader' OR date_from < '1989-04-24 23:51:54'
Qual é a última data em que o pessoal deixa os projetos?
CREATE TABLE Project_Staff (date_to VARCHAR)
SELECT date_to FROM Project_Staff ORDER BY date_to DESC LIMIT 1
Qual é a descrição do resultado do projecto cujo detalhe é "sint"?
CREATE TABLE Project_outcomes (outcome_code VARCHAR, project_id VARCHAR); CREATE TABLE Research_outcomes (outcome_description VARCHAR, outcome_code VARCHAR); CREATE TABLE Projects (project_id VARCHAR, project_details VARCHAR)
SELECT T1.outcome_description FROM Research_outcomes AS T1 JOIN Project_outcomes AS T2 ON T1.outcome_code = T2.outcome_code JOIN Projects AS T3 ON T2.project_id = T3.project_id WHERE T3.project_details = 'sint'
Indicar o ID da organização com a contagem máxima de resultados e a contagem.
CREATE TABLE Project_outcomes (project_id VARCHAR); CREATE TABLE Projects (organisation_id VARCHAR, project_id VARCHAR)
SELECT T1.organisation_id, COUNT(*) FROM Projects AS T1 JOIN Project_outcomes AS T2 ON T1.project_id = T2.project_id GROUP BY T1.organisation_id ORDER BY COUNT(*) DESC LIMIT 1
Lista dos pormenores dos projectos lançados pela organização
CREATE TABLE Projects (project_details VARCHAR, organisation_id VARCHAR)
SELECT project_details FROM Projects WHERE organisation_id IN (SELECT organisation_id FROM Projects GROUP BY organisation_id ORDER BY COUNT(*) DESC LIMIT 1)
Escreva os detalhes do pessoal de investigação, e em ordem ascendente.
CREATE TABLE Research_Staff (staff_details VARCHAR)
SELECT staff_details FROM Research_Staff ORDER BY staff_details
Quantas tarefas existem no total?
CREATE TABLE Tasks (Id VARCHAR)
SELECT COUNT(*) FROM Tasks
Quantas tarefas tem cada projeto?
CREATE TABLE Projects (project_details VARCHAR, project_id VARCHAR); CREATE TABLE Tasks (project_id VARCHAR)
SELECT COUNT(*), T1.project_details FROM Projects AS T1 JOIN Tasks AS T2 ON T1.project_id = T2.project_id GROUP BY T1.project_id
Quais são os papéis do pessoal que
CREATE TABLE Project_Staff (role_code VARCHAR, date_from VARCHAR, date_to VARCHAR)
SELECT role_code FROM Project_Staff WHERE date_from > '2003-04-19 15:06:20' AND date_to < '2016-03-15 00:33:18'
Quais são as descrições de todos os resultados do projecto?
CREATE TABLE Research_outcomes (outcome_description VARCHAR, outcome_code VARCHAR); CREATE TABLE Project_outcomes (outcome_code VARCHAR)
SELECT T1.outcome_description FROM Research_outcomes AS T1 JOIN Project_outcomes AS T2 ON T1.outcome_code = T2.outcome_code
Qual é a função mais comum para o pessoal?
CREATE TABLE Project_Staff (role_code VARCHAR)
SELECT role_code FROM Project_Staff GROUP BY role_code ORDER BY COUNT(*) DESC LIMIT 1
Quantos amigos tem o Dan?
CREATE TABLE PersonFriend (friend VARCHAR, name VARCHAR); CREATE TABLE Person (name VARCHAR)
SELECT COUNT(T2.friend) FROM Person AS T1 JOIN PersonFriend AS T2 ON T1.name = T2.name WHERE T1.name = 'Dan'
Quantas mulheres tem esta rede?
CREATE TABLE Person (gender VARCHAR)
SELECT COUNT(*) FROM Person WHERE gender = 'female'
Qual é a idade média de todas as pessoas?
CREATE TABLE Person (age INTEGER)
SELECT AVG(age) FROM Person
De quantas cidades diferentes são?
CREATE TABLE Person (city VARCHAR)
SELECT COUNT(DISTINCT city) FROM Person
Quantos tipos de empregos eles têm?
CREATE TABLE Person (job VARCHAR)
SELECT COUNT(DISTINCT job) FROM Person
Quem é a pessoa mais velha?
CREATE TABLE Person (name VARCHAR, age INTEGER); CREATE TABLE person (name VARCHAR, age INTEGER)
SELECT name FROM Person WHERE age = (SELECT MAX(age) FROM person)
Quem é a pessoa mais velha que trabalha como estudante?
CREATE TABLE person (name VARCHAR, job VARCHAR, age INTEGER); CREATE TABLE Person (name VARCHAR, job VARCHAR, age INTEGER)
SELECT name FROM Person WHERE job = 'student' AND age = (SELECT MAX(age) FROM person WHERE job = 'student')
Quem é o mais novo?
CREATE TABLE Person (name VARCHAR, gender VARCHAR, age INTEGER); CREATE TABLE person (name VARCHAR, gender VARCHAR, age INTEGER)
SELECT name FROM Person WHERE gender = 'male' AND age = (SELECT MIN(age) FROM person WHERE gender = 'male')
Quantos anos tem o médico chamado Zach?
CREATE TABLE Person (age VARCHAR, job VARCHAR, name VARCHAR)
SELECT age FROM Person WHERE job = 'doctor' AND name = 'Zach'
Quem é a pessoa cuja idade é inferior a 30 anos?
CREATE TABLE Person (name VARCHAR, age INTEGER)
SELECT name FROM Person WHERE age < 30
Quantas pessoas com mais de 30 anos e trabalho são engenheiros?
CREATE TABLE Person (age VARCHAR, job VARCHAR)
SELECT COUNT(*) FROM Person WHERE age > 30 AND job = 'engineer'
Qual é a idade média para cada sexo?
CREATE TABLE Person (gender VARCHAR, age INTEGER)
SELECT AVG(age), gender FROM Person GROUP BY gender
Qual é a idade média para diferentes títulos de trabalho?
CREATE TABLE Person (job VARCHAR, age INTEGER)
SELECT AVG(age), job FROM Person GROUP BY job
Qual é a idade média dos homens para diferentes cargos?
CREATE TABLE Person (job VARCHAR, age INTEGER, gender VARCHAR)
SELECT AVG(age), job FROM Person WHERE gender = 'male' GROUP BY job
Qual é a idade mínima para diferentes títulos de emprego?
CREATE TABLE Person (job VARCHAR, age INTEGER)
SELECT MIN(age), job FROM Person GROUP BY job
Encontre o número de pessoas com menos de 40 anos para cada sexo.
CREATE TABLE Person (gender VARCHAR, age INTEGER)
SELECT COUNT(*), gender FROM Person WHERE age < 40 GROUP BY gender
Encontre o nome de pessoas cuja idade é maior do que qualquer engenheiro, ordenado por idade.
CREATE TABLE Person (name VARCHAR, age INTEGER, job VARCHAR); CREATE TABLE person (name VARCHAR, age INTEGER, job VARCHAR)
SELECT name FROM Person WHERE age > (SELECT MIN(age) FROM person WHERE job = 'engineer') ORDER BY age
Encontre o número de pessoas cuja idade é maior do que todos os engenheiros.
CREATE TABLE Person (age INTEGER, job VARCHAR); CREATE TABLE person (age INTEGER, job VARCHAR)
SELECT COUNT(*) FROM Person WHERE age > (SELECT MAX(age) FROM person WHERE job = 'engineer')
Escrever o nome, o cargo de todas as pessoas ordenadas por seus nomes.
CREATE TABLE Person (name VARCHAR, job VARCHAR)
SELECT name, job FROM Person ORDER BY name
Encontre os nomes de todas as pessoas ordenadas em ordem decrescente usando a idade.
CREATE TABLE Person (name VARCHAR, age VARCHAR)
SELECT name FROM Person ORDER BY age DESC
Encontre o nome e a idade de todos os machos em ordem de idade.
CREATE TABLE Person (name VARCHAR, gender VARCHAR, age VARCHAR)
SELECT name FROM Person WHERE gender = 'male' ORDER BY age
Encontre o nome e a idade da pessoa que é amiga de Dan e Alice.
CREATE TABLE Person (name VARCHAR, age VARCHAR); CREATE TABLE PersonFriend (name VARCHAR, friend VARCHAR)
SELECT T1.name, T1.age FROM Person AS T1 JOIN PersonFriend AS T2 ON T1.name = T2.name WHERE T2.friend = 'Dan' INTERSECT SELECT T1.name, T1.age FROM Person AS T1 JOIN PersonFriend AS T2 ON T1.name = T2.name WHERE T2.friend = 'Alice'
Encontre o nome e a idade da pessoa que é amiga de Dan ou Alice.
CREATE TABLE Person (name VARCHAR, age VARCHAR); CREATE TABLE PersonFriend (name VARCHAR, friend VARCHAR)
SELECT DISTINCT T1.name, T1.age FROM Person AS T1 JOIN PersonFriend AS T2 ON T1.name = T2.name WHERE T2.friend = 'Dan' OR T2.friend = 'Alice'
Encontre o nome da pessoa que tem amigos com idade acima de 40 e abaixo de 30 anos?
CREATE TABLE Person (name VARCHAR, age INTEGER); CREATE TABLE PersonFriend (name VARCHAR, friend VARCHAR); CREATE TABLE Person (name VARCHAR)
SELECT T1.name FROM Person AS T1 JOIN PersonFriend AS T2 ON T1.name = T2.name WHERE T2.friend IN (SELECT name FROM Person WHERE age > 40) INTERSECT SELECT T1.name FROM Person AS T1 JOIN PersonFriend AS T2 ON T1.name = T2.name WHERE T2.friend IN (SELECT name FROM Person WHERE age < 30)
Encontre o nome da pessoa que tem amigos com mais de 40 anos, mas não com menos de 30?
CREATE TABLE Person (name VARCHAR, age INTEGER); CREATE TABLE PersonFriend (name VARCHAR, friend VARCHAR); CREATE TABLE Person (name VARCHAR)
SELECT T1.name FROM Person AS T1 JOIN PersonFriend AS T2 ON T1.name = T2.name WHERE T2.friend IN (SELECT name FROM Person WHERE age > 40) EXCEPT SELECT T1.name FROM Person AS T1 JOIN PersonFriend AS T2 ON T1.name = T2.name WHERE T2.friend IN (SELECT name FROM Person WHERE age < 30)
Encontre o nome da pessoa que não tem amigos estudantes.
CREATE TABLE Person (name VARCHAR, job VARCHAR); CREATE TABLE PersonFriend (name VARCHAR, friend VARCHAR); CREATE TABLE person (name VARCHAR)
SELECT name FROM person EXCEPT SELECT T2.name FROM Person AS T1 JOIN PersonFriend AS T2 ON T1.name = T2.friend WHERE T1.job = 'student'
Encontre a pessoa que tem exatamente um amigo.
CREATE TABLE PersonFriend (name VARCHAR)
SELECT name FROM PersonFriend GROUP BY name HAVING COUNT(*) = 1
Quem são os amigos do Bob?
CREATE TABLE PersonFriend (friend VARCHAR, name VARCHAR); CREATE TABLE Person (name VARCHAR)
SELECT T2.friend FROM Person AS T1 JOIN PersonFriend AS T2 ON T1.name = T2.name WHERE T1.name = 'Bob'
Encontre o nome das pessoas que são amigas do Bob.
CREATE TABLE PersonFriend (name VARCHAR, friend VARCHAR); CREATE TABLE Person (name VARCHAR)
SELECT T1.name FROM Person AS T1 JOIN PersonFriend AS T2 ON T1.name = T2.name WHERE T2.friend = 'Bob'
Encontre os nomes das mulheres que são amigas do Zach.
CREATE TABLE PersonFriend (name VARCHAR, friend VARCHAR); CREATE TABLE Person (name VARCHAR, gender VARCHAR)
SELECT T1.name FROM Person AS T1 JOIN PersonFriend AS T2 ON T1.name = T2.name WHERE T2.friend = 'Zach' AND T1.gender = 'female'
Encontrem as amigas da Alice.
CREATE TABLE PersonFriend (friend VARCHAR, name VARCHAR); CREATE TABLE Person (name VARCHAR, gender VARCHAR)
SELECT T2.friend FROM Person AS T1 JOIN PersonFriend AS T2 ON T1.name = T2.friend WHERE T2.name = 'Alice' AND T1.gender = 'female'
Encontrar o amigo do homem da Alice que é médico?
CREATE TABLE PersonFriend (friend VARCHAR, name VARCHAR); CREATE TABLE Person (name VARCHAR, job VARCHAR, gender VARCHAR)
SELECT T2.friend FROM Person AS T1 JOIN PersonFriend AS T2 ON T1.name = T2.friend WHERE T2.name = 'Alice' AND T1.gender = 'male' AND T1.job = 'doctor'
Quem tem um amigo que é de Nova Iorque?
CREATE TABLE Person (name VARCHAR, city VARCHAR); CREATE TABLE PersonFriend (name VARCHAR, friend VARCHAR)
SELECT T2.name FROM Person AS T1 JOIN PersonFriend AS T2 ON T1.name = T2.friend WHERE T1.city = 'new york city'
Quem tem amigos que são mais jovens do que a idade média?
CREATE TABLE person (age INTEGER); CREATE TABLE Person (name VARCHAR, age INTEGER); CREATE TABLE PersonFriend (name VARCHAR, friend VARCHAR)
SELECT DISTINCT T2.name FROM Person AS T1 JOIN PersonFriend AS T2 ON T1.name = T2.friend WHERE T1.age < (SELECT AVG(age) FROM person)
Quem tem amigos que são mais velhos do que a idade média?
CREATE TABLE person (age INTEGER); CREATE TABLE PersonFriend (name VARCHAR, friend VARCHAR); CREATE TABLE Person (age INTEGER, name VARCHAR)
SELECT DISTINCT T2.name, T2.friend, T1.age FROM Person AS T1 JOIN PersonFriend AS T2 ON T1.name = T2.friend WHERE T1.age > (SELECT AVG(age) FROM person)
Quem é o amigo do Zach com o relacionamento mais longo?
CREATE TABLE PersonFriend (friend VARCHAR, name VARCHAR, YEAR INTEGER)
SELECT friend FROM PersonFriend WHERE name = 'Zach' AND YEAR = (SELECT MAX(YEAR) FROM PersonFriend WHERE name = 'Zach')
Qual é a idade do amigo do Zach com o relacionamento mais longo?
CREATE TABLE PersonFriend (friend VARCHAR, name VARCHAR, year VARCHAR); CREATE TABLE PersonFriend (YEAR INTEGER, name VARCHAR); CREATE TABLE Person (age VARCHAR, name VARCHAR)
SELECT T1.age FROM Person AS T1 JOIN PersonFriend AS T2 ON T1.name = T2.friend WHERE T2.name = 'Zach' AND T2.year = (SELECT MAX(YEAR) FROM PersonFriend WHERE name = 'Zach')
Encontre o nome das pessoas que são amigos de Alice para os anos mais curtos.
CREATE TABLE PersonFriend (name VARCHAR, friend VARCHAR, YEAR INTEGER)
SELECT name FROM PersonFriend WHERE friend = 'Alice' AND YEAR = (SELECT MIN(YEAR) FROM PersonFriend WHERE friend = 'Alice')
Encontre o nome, idade e cargo de pessoas que são amigos de Alice há mais anos.
CREATE TABLE PersonFriend (name VARCHAR, friend VARCHAR, year VARCHAR); CREATE TABLE Person (name VARCHAR, age VARCHAR, job VARCHAR); CREATE TABLE PersonFriend (YEAR INTEGER, friend VARCHAR)
SELECT T1.name, T1.age, T1.job FROM Person AS T1 JOIN PersonFriend AS T2 ON T1.name = T2.name WHERE T2.friend = 'Alice' AND T2.year = (SELECT MAX(YEAR) FROM PersonFriend WHERE friend = 'Alice')
Quem é a pessoa que não tem um amigo?
CREATE TABLE PersonFriend (name VARCHAR); CREATE TABLE person (name VARCHAR)
SELECT name FROM person EXCEPT SELECT name FROM PersonFriend
Qual pessoa cujos amigos têm a idade média mais avançada?
CREATE TABLE PersonFriend (name VARCHAR, friend VARCHAR); CREATE TABLE Person (age INTEGER, name VARCHAR)
SELECT T2.name, AVG(T1.age) FROM Person AS T1 JOIN PersonFriend AS T2 ON T1.name = T2.friend GROUP BY T2.name ORDER BY AVG(T1.age) DESC LIMIT 1
Qual é o número total de pessoas que não tem amigos vivendo na cidade de Austin.
CREATE TABLE person (name VARCHAR, friend VARCHAR, city VARCHAR); CREATE TABLE PersonFriend (name VARCHAR, friend VARCHAR, city VARCHAR)
SELECT COUNT(DISTINCT name) FROM PersonFriend WHERE NOT friend IN (SELECT name FROM person WHERE city = 'Austin')
Encontrem os amigos de amigos da Alice.
CREATE TABLE PersonFriend (name VARCHAR); CREATE TABLE PersonFriend (name VARCHAR, friend VARCHAR); CREATE TABLE Person (name VARCHAR)
SELECT DISTINCT T4.name FROM PersonFriend AS T1 JOIN Person AS T2 ON T1.name = T2.name JOIN PersonFriend AS T3 ON T1.friend = T3.name JOIN PersonFriend AS T4 ON T3.friend = T4.name WHERE T2.name = 'Alice' AND T4.name <> 'Alice'
Quantos membros há?
CREATE TABLE member (Id VARCHAR)
SELECT COUNT(*) FROM member
Escreva os nomes dos membros em ordem alfabética ascendente.
CREATE TABLE member (Name VARCHAR)
SELECT Name FROM member ORDER BY Name
Quais são os nomes e países dos membros?
CREATE TABLE member (Name VARCHAR, Country VARCHAR)
SELECT Name, Country FROM member
Mostre os nomes dos membros cujo país é "Estados Unidos" ou "Canadá".
CREATE TABLE member (Name VARCHAR, Country VARCHAR)
SELECT Name FROM member WHERE Country = "United States" OR Country = "Canada"
Indicar os diferentes países e o número de membros de cada um.
CREATE TABLE member (Country VARCHAR)
SELECT Country, COUNT(*) FROM member GROUP BY Country
Mostre o país mais comum entre os membros.
CREATE TABLE member (Country VARCHAR)
SELECT Country FROM member GROUP BY Country ORDER BY COUNT(*) DESC LIMIT 1
Que países têm mais de dois membros?
CREATE TABLE member (Country VARCHAR)
SELECT Country FROM member GROUP BY Country HAVING COUNT(*) > 2
Mostre aos líderes os nomes e localizações das faculdades.
CREATE TABLE college (Leader_Name VARCHAR, College_Location VARCHAR)
SELECT Leader_Name, College_Location FROM college
Mostrar os nomes dos membros e os nomes das faculdades que eles vão.
CREATE TABLE member (Name VARCHAR, College_ID VARCHAR); CREATE TABLE college (Name VARCHAR, College_ID VARCHAR)
SELECT T2.Name, T1.Name FROM college AS T1 JOIN member AS T2 ON T1.College_ID = T2.College_ID
Mostrar os nomes dos membros e os locais das faculdades que eles vão em ordem alfabética ascendente dos nomes dos membros.
CREATE TABLE member (Name VARCHAR, College_ID VARCHAR); CREATE TABLE college (College_Location VARCHAR, College_ID VARCHAR)
SELECT T2.Name, T1.College_Location FROM college AS T1 JOIN member AS T2 ON T1.College_ID = T2.College_ID ORDER BY T2.Name
Mostrar os nomes dos principais colégios associados com membros do país "Canadá".
CREATE TABLE college (Leader_Name VARCHAR, College_ID VARCHAR); CREATE TABLE member (College_ID VARCHAR, Country VARCHAR)
SELECT DISTINCT T1.Leader_Name FROM college AS T1 JOIN member AS T2 ON T1.College_ID = T2.College_ID WHERE T2.Country = "Canada"
Mostrar os nomes dos membros e os temas de decoração que eles têm.
CREATE TABLE member (Name VARCHAR, Member_ID VARCHAR); CREATE TABLE round (Decoration_Theme VARCHAR, Member_ID VARCHAR)
SELECT T1.Name, T2.Decoration_Theme FROM member AS T1 JOIN round AS T2 ON T1.Member_ID = T2.Member_ID
Mostrar os nomes dos membros que têm um ranking em rodada superior a 3.
CREATE TABLE round (Member_ID VARCHAR, Rank_in_Round INTEGER); CREATE TABLE member (Name VARCHAR, Member_ID VARCHAR)
SELECT T1.Name FROM member AS T1 JOIN round AS T2 ON T1.Member_ID = T2.Member_ID WHERE T2.Rank_in_Round > 3
Mostrar os nomes dos membros em ordem ascendente de sua patente em rondas.
CREATE TABLE member (Name VARCHAR, Member_ID VARCHAR); CREATE TABLE round (Member_ID VARCHAR)
SELECT T1.Name FROM member AS T1 JOIN round AS T2 ON T1.Member_ID = T2.Member_ID ORDER BY Rank_in_Round
Escreva os nomes dos membros que não participaram em nenhuma rodada.
CREATE TABLE member (Name VARCHAR, Member_ID VARCHAR); CREATE TABLE round (Name VARCHAR, Member_ID VARCHAR)
SELECT Name FROM member WHERE NOT Member_ID IN (SELECT Member_ID FROM round)
Encontre o nome e o número de acessos de todos os documentos, em ordem alfabética do nome do documento.
CREATE TABLE documents (document_name VARCHAR, access_count VARCHAR)
SELECT document_name, access_count FROM documents ORDER BY document_name
Encontre o nome do documento que foi acessado o maior número de vezes, bem como a contagem de quantas vezes ele foi acessado?
CREATE TABLE documents (document_name VARCHAR, access_count VARCHAR)
SELECT document_name, access_count FROM documents ORDER BY access_count DESC LIMIT 1
Encontrar os tipos de documentos com mais de 4 documentos.
CREATE TABLE documents (document_type_code VARCHAR)
SELECT document_type_code FROM documents GROUP BY document_type_code HAVING COUNT(*) > 4
Encontre a contagem total de acessos de todos os documentos no tipo de documento mais popular.
CREATE TABLE documents (access_count INTEGER, document_type_code VARCHAR)
SELECT SUM(access_count) FROM documents GROUP BY document_type_code ORDER BY COUNT(*) DESC LIMIT 1
Qual é a média de documentos que são objeto de acesso?
CREATE TABLE documents (access_count INTEGER)
SELECT AVG(access_count) FROM documents
Qual é a estrutura do documento com o menor número de acessos?
CREATE TABLE document_structures (document_structure_description VARCHAR, document_structure_code VARCHAR); CREATE TABLE documents (document_structure_code VARCHAR)
SELECT t2.document_structure_description FROM documents AS t1 JOIN document_structures AS t2 ON t1.document_structure_code = t2.document_structure_code GROUP BY t1.document_structure_code ORDER BY COUNT(*) DESC LIMIT 1
Qual é o tipo de documento chamado "CV de David"?
CREATE TABLE documents (document_type_code VARCHAR, document_name VARCHAR)
SELECT document_type_code FROM documents WHERE document_name = "David CV"
Encontre a lista de documentos que estão ambos no tipo mais popular e têm a estrutura mais popular.
CREATE TABLE documents (document_name VARCHAR, document_type_code VARCHAR, document_structure_code VARCHAR)
SELECT document_name FROM documents GROUP BY document_type_code ORDER BY COUNT(*) DESC LIMIT 3 INTERSECT SELECT document_name FROM documents GROUP BY document_structure_code ORDER BY COUNT(*) DESC LIMIT 3
Quais tipos de documentos têm mais de 10000 números de acesso totais?
CREATE TABLE documents (document_type_code VARCHAR, access_count INTEGER)
SELECT document_type_code FROM documents GROUP BY document_type_code HAVING SUM(access_count) > 10000
Quais são os títulos das secções do documento chamado "CV de David"?
CREATE TABLE documents (document_code VARCHAR, document_name VARCHAR); CREATE TABLE document_sections (section_title VARCHAR, document_code VARCHAR)
SELECT t2.section_title FROM documents AS t1 JOIN document_sections AS t2 ON t1.document_code = t2.document_code WHERE t1.document_name = "David CV"