Question

ACTOR (id, fname, lname, gender)
MOVIE (id, name, year, rank)
CASTS (pid, mid, role)
WHERE pid references ACTOR id
mid references Movie id

列出x没有y的电影（x和y是演员）。

我发现很难用NOT编译SQL。这是我的尝试。由于第二个演员没有出现，我无法完成它

SELECT m.name
FROM MOVIE m
WHERE m.id NOT IN (SELECT c.mid
                       FROM CASTS c, ACTOR a
                       WHERE c.pid = a.id AND a.name = "adam..")

Answer 1

使用 NOT EXISTS ：

SELECT m.name                       -- Show the names                      
FROM movie m                        -- of all movies
WHERE EXISTS                        -- that there was
      ( SELECT *                    -- a role
        FROM casts c                -- casted to
          JOIN actor a              -- actor with
            ON c.pid = a.id
        WHERE c.mid = m.id  
          AND a.name = 'Actor X'    -- name X
      ) 
  AND NOT EXISTS                    -- and there was not
      ( SELECT *                    -- any role
        FROM casts c                -- casted
          JOIN actor a              -- to actor with
            ON c.pid = a.id
        WHERE c.mid = m.id 
          AND a.name = 'Actor Y'    -- name Y
      ) ;

您还可以使用 NOT IN 。请注意，如果NULL或movie.id列中有casts.mid行，则可能会给您带来意外结果：

SELECT m.name                       -- Show the names                      
FROM movie m                        -- of all movies
WHERE m.id IN                       -- but keep only the movies that
      ( SELECT c.mid                -- movies that
        FROM casts c                -- had a role casted to
          JOIN actor a              -- actor with
            ON c.pid = a.id
        WHERE a.name = 'Actor X'    -- name X
      ) 
  AND m.id NOT IN                   -- and not the movies
      ( SELECT c.mid                -- that
        FROM casts c                -- had a role casted
          JOIN actor a              -- to actor with
            ON c.pid = a.id
        WHERE a.name = 'Actor Y'    -- name Y
      ) ;

Answer 2

您还可以使用经常被忽视的MINUS：

SELECT Movie.id, Movie.name
  FROM Actor
  INNER JOIN Casts ON Actor.id = Casts.pid
  INNER JOIN Movie ON Casts.mid = Movie.id
  WHERE Actor.id = 1
MINUS SELECT Movie.id, Movie.name
  FROM Actor
  INNER JOIN Casts ON Actor.id = Casts.pid
  INNER JOIN Movie ON Casts.mid = Movie.id
  WHERE Actor.id = 2

上述查询中的WHERE Actor.id可以用其他方式替换，以唯一地标识actor，例如通过其名称。

Answer 3

SELECT  a.*
FROM    Movie a
        INNER JOIN Casts b
            ON a.ID = b.mID
        INNER JOIN Actor c
            ON b.pid = c.ID
        LEFT JOIN
        (
            SELECT  aa.mid
            FROM    Casts aa
                    INNER JOIN Actor bb
                        ON aa.pid = bb.ID
            WHERE   bb.fName = 'Y_Name'
        ) d ON  a.id = d.mid
WHERE   c.fname = 'X_Name' AND
        d.mid IS NULL

SQLFiddle Demo

在子查询上进行额外连接的原因是因为我们按照actor的名称过滤记录。

假设你有这些记录集

ACTOR

╔════╦════════╗
║ ID ║ FNAME  ║
╠════╬════════╣
║  1 ║ X_Name ║
║  2 ║ Y_Name ║
╚════╩════════╝

MOVIE

╔════╦══════╗
║ ID ║ NAME ║
╠════╬══════╣
║  1 ║ Mov1 ║
║  2 ║ Mov2 ║
║  3 ║ Mov3 ║
╚════╩══════╝

CAST

╔═════╦═════╗
║ PID ║ MID ║
╠═════╬═════╣
║   1 ║   1 ║ <<== EXPECTED OUTPUT since Y_NAME is not present
║   1 ║   2 ║                      on Movie Mov1
║   2 ║   2 ║
║   1 ║   3 ║
║   2 ║   3 ║
╚═════╩═════╝

输出

╔════╦══════╗
║ ID ║ NAME ║
╠════╬══════╣
║  1 ║ Mov1 ║
╚════╩══════╝

Answer 4

首先获取Actor X的所有电影。然后检查过滤掉任何还包含Actor Y的电影。

SELECT m.name
FROM MOVIE m, CASTS c, ACTOR a
WHERE m.id = c.mid
    AND c.pid = a.id
    AND a.name = "ACTOR X"
    AND NOT EXISTS (
        SELECT 1
        FROM CASTS c1, ACTOR a1
        WHERE c1.pid = a1.id
            AND m.id = c1.mid
            AND a1.name = "ACTOR Y"
    )

不使用子查询SQL构造

4 个答案: