Question

此代码未按预期运行。我想要一起或单独选择所有包含TAGS 561, 562的帖子，但没有任何此处不存在的标记 - ＆gt;＆gt;＆gt; IN (561, 562)

 SELECT post_id_post
 FROM post_has_tags
 WHERE tags_id_tags IN (561, 562)
 HAVING COUNT(tags_id_tags) <= 2 

post_id_post        tags_id_tags    
600                    561  
600                    562  
600                    917  // 917 is not inside IN (561, 562)

但是这段代码会输出帖子600.这是错误的，正确的输出应该没有结果。

感谢

Answer 1

试试这个：

select distinct p1.post_id_post from post_has_tags p1
where not exists (
  select * from post_has_tags p2
  where p1.post_id_post = p2.post_id_post and p2.tags_id_tags not in (561, 562)
)

我不想小于菲尔，所以我也在添加我的fiddle：）

Answer 2

SELECT DISTINCT pht.post_id_post -- the distinct is to avoid duplicates
FROM post_has_tags pht
WHERE pht.tags_id_tags IN (561, 562)
AND NOT EXISTS (
    SELECT 1 FROM post_has_tags _pht
    WHERE _pht.post_id_post = pht.post_id_post
    AND _pht.tags_id_tags NOT IN (561, 562)
);

在这里演示 - http://sqlfiddle.com/#!2/cdef3/3

Answer 3

可能有点矫枉过正，但是：

SELECT post_id_post
FROM post_has_tags
WHERE tags_id_tags IN (561, 562)
AND tags_id_tags NOT IN (SELECT DISTINCT(tag_id) FROM tags WHERE tag NOT IN (561, 562));

（假设你有一个单独的标签表）

得到错误的选择

3 个答案: