说有这样的表:
mysql> SELECT * FROM tags;
+---------+--------+
| post_id | tag_id |
+---------+--------+
| 1 | 2 |
| 1 | 3 |
| 1 | 1 |
| 2 | 1 |
| 2 | 2 |
+---------+--------+
5 rows in set (0.00 sec)
字段名称非常明显。我想选择同时包含1和3 post_id
s的tag_id
,因此在此示例中,它仅为1
。我想到了类似的东西
SELECT post_id FROM tags GROUP BY post_id HAVING ...
在我想列出此群组中的tag_id
之后。我该怎么做?
答案 0 :(得分:6)
如果post_id和tag_id都有唯一约束,那么它也应该有效:
SELECT post_id
FROM tags
WHERE tag_id = 1 OR tag_id = 3
GROUP BY post_id
HAVING count(*) = 2;
如果没有任何唯一约束,请尝试:
SELECT DISTINCT post_id
FROM tags
WHERE tag_id = 1 OR tag_id = 3
GROUP BY post_id
HAVING count(DISTINCT tag_id) = 2;
答案 1 :(得分:2)
您可以尝试自我加入(N tag_id - > N加入),但可能不是很快
SELECT t1.post_id
FROM tags t1 INNER JOIN tags t2 ON t1.post_id = t2.post_id
WHERE t1.tag_id = 1 AND t2.tag_id = 3
答案 2 :(得分:1)
我对你的其他表做了一些假设。 (即你有一个我称为posts
的帖子的表和一个我用tag_table
作为PK的标签的表来避免使用帖子/标签表的名字标记,我已经可以看到你了致电tags
)
您希望列表{1,3}中不存在标记的帖子与相应的post_id / tag_id不存在匹配记录,因此您可以使用双重NOT EXISTS结构,如下所示。
SELECT post_id
FROM posts p
WHERE NOT EXISTS
(SELECT * FROM tag_table tt
WHERE tag_id IN (1,3)
AND NOT EXISTS
(SELECT * FROM tags t
WHERE t.tag_id = tt.tag_id and
p.post_id = t.post_id)
)
另一种替代方法是使用Group By和Count。 A review of approaches to this problem is here
答案 3 :(得分:1)
SELECT post_id
FROM ( SELECT post_id,
count(tag_id) AS counter
FROM tags
WHERE tag_id IN (1,3)
GROUP BY post_id
)
WHERE counter = 2
在问题的第二部分使用GROUP_CONCAT()
SELECT post_id,
GROUP_CONCAT(tag_id ORDER BY tag_id ASC SEPARATOR ',')
FROM tags
答案 4 :(得分:0)
怎么样
SELECT *
FROM tags
WHERE post_id in
(SELECT post_id AS pid
FROM tags
WHERE 1 IN (SELECT tag_id FROM tags WHERE post_id = pid)
AND 3 IN (SELECT tag_id FROM tags WHERE post_id = pid)
);
答案 5 :(得分:0)
@ Keeper解决方案的版本
SELECT DISTINCT t1.post_id
FROM tags t1, tags t2
WHERE
t1.post_id = t2.post_id AND
t1.tag_id = 1 AND t2.tag_id = 3