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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" |
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