複雜的SQL查詢多對多

Question

我已經創建了數據庫這樣的：

我想選擇的人，電影的名字是同一個人充當「Oyuncu」和「Senarist」

請幫幫我。我寫了一個這樣的查詢;

select 
    MovieName,PersonName,RoleName 
from 
    Movies, 
    MoviesPersonRole, 
    PersonRole, 
    Person, 
    Role 
where 
    Movies.Id = MoviesPersonRole.MovieId 
    and 
    PersonRole.Id = MoviesPersonRole.PersonRoleId 
    and 
    PersonRole.RoleId = Role.Id 
    and 
    PersonRole.PersonId = Person.Id 

，結果

MovieName   PersonName RoleName 
Pulp Fiction  Mehmet  Oyuncu 
Pulp Fiction  Mehmet  Senarist 
Matrix    Aylin   Oyuncu 
LOTR    Gökberk  Oyuncu 
LOTR    Gökberk  Senarist 
Pulp Fiction  Aylin   Oyuncu 

Answer 1

你可以嘗試somethong這樣的：

SELECT X.MovieName, X.PersonName FROM (
    SELECT M.MovieName, PE.PersonName, COUNT(*) AS RoleCount 
    FROM Movies M 
    INNER JOIN MoviesPersonRole MPR ON (MPR.MovieId=M.Id) 
    INNER JOIN PersonRole PR ON (PR.Id = MPR.PersonRoleId) 
    INNER JOIN Person PE ON (PE.Id = PR.PersonId) 
    INNER JOIN Role R ON (R.Id = PR.RoleId) 
    WHERE R.RoleName='Oyuncu' OR R.RoleName='Senarist' 
    GROUP BY M.MovieName, PE.PersonName 
) X WHERE X.RoleCount>1 

Answer 2

加入這一行查詢

and Role.RoleName in('Oyuncu' ,'Senarist') 

Answer 3

第二個加入的角色應該這樣做：

SELECT MovieName , 
     PersonName , 
     RoleName 
FROM Movies 
     JOIN MoviesPersonRole ON Movies.Id = MoviesPersonRole.MovieId 
     JOIN PersonRole ON PersonRole.Id = MoviesPersonRole.PersonRoleId 
     JOIN Person ON PersonRole.PersonId = Person.Id 
     JOIN Role Role1 ON PersonRole.RoleId = Role1.Id 
     JOIN Role Role2 ON PersonRole.RoleId = Role2.Id 
WHERE Role1.RoleName = 'Oyuncu' 
     AND Role2.RoleName = 'Senarist' 

Answer 4

以下查詢嘗試：

SELECT MovieName , 
     PersonName , 
     RoleName 
FROM Movies, 
     MoviesPersonRole, 
     PersonRole, 
     Person, 
     Role 
WHERE Role1.RoleName = 'Oyuncu' 
     AND Role2.RoleName = 'Senarist' 
     AND Movies.Id = MoviesPersonRole.MovieId 
     AND PersonRole.Id = MoviesPersonRole.PersonRoleId 
     AND PersonRole.RoleId = Role.Id 
     AND PersonRole.PersonId = Person.Id