所以我有一个“流派”表,它有重复的流派和一个id,我想制定一个查询,以便我返回所有具有特定id的流派的行。
例如,
+---------+-------+
| genre | movie |
+---------+-------+
| Musical | 558 |
| Musical | 562 |
| Musical | 597 |
| Musical | 651 |
| Musical | 656 |
| Musical | 791 |
| Musical | 810 |
| Musical | 845 |
| Musical | 859 |
| Musical | 919 |
| Musical | 949 |
| Musical | 971 |
+---------+-------+
12 rows in set (0.00 sec)
和
+--------+-------+
| genre | movie |
+--------+-------+
| Comedy | 642 |
| Comedy | 643 |
| Comedy | 644 |
| Comedy | 651 |
| Comedy | 654 |
| Comedy | 658 |
+--------+-------+
6 rows in set (0.00 sec)
我希望它能回归“电影651”,因为它既有喜剧也有音乐(假设对“喜剧”和“音乐剧”都有疑问)。
答案 0 :(得分:4)
使用
SELECT DISTINCT A.MOVIE FROM GENRES A, GENRES B WHERE A.MOVIE = B.MOVIE AND A.GENRE = 'Comedy' AND B.GENRE = 'Musical';
编辑 - 根据评论:
SELECT DISTINCT A.MOVIE
FROM
GENRES A
INNER JOIN
GENRES B
ON A.MOVIE = B.MOVIE
WHERE A.GENRE = 'Comedy'
AND B.GENRE = 'Musical';
答案 1 :(得分:1)
SELECT DISTINCT
movie
FROM
genres
WHERE
movie in (SELECT movie from genres where genre = "Musical")
AND movie in (SELECT movie from genres where genre = "Comedy")
答案 2 :(得分:0)
这将为您提供
中的ID列表select movie from table a join table b on a.movie = b.movie