带COUNT的LEFT JOIN返回意外值

时间:2016-04-11 00:24:00

标签: mysql join count left-join inner-join

我有两张桌子:

发表

id | body   | author | type | date
1  | hi!    | Igor   | 2    | 04-10
2  | hello! | Igor   | 1    | 04-10
3  | lol    | Igor   | 1    | 04-10
4  | good!  | Igor   | 3    | 04-10
5  | nice!  | Igor   | 2    | 04-10
6  | count  | Igor   | 3    | 04-10
7  | left   | Igor   | 3    | 04-10
8  | join   | Igor   | 4    | 04-10

喜欢

id | author | post_id
1  | Igor   | 2
2  | Igor   | 5
3  | Igor   | 6
4  | Igor   | 8

我想做一个查询,返回Igor用2型,3型或4型制作的帖子数量以及Igor制作的喜欢的数量,所以,我做了:

SELECT COUNT(DISTINCT p.type = 2 OR p.type = 3 OR p.type = 4) AS numberPhotos, COUNT(DISTINCT l.id) AS numberLikes
FROM post p
LEFT JOIN likes l
ON p.author = l.author
WHERE p.author = 'Igor'

预期的输出是:

array(1) {
  [0]=>
  array(2) {
    ["numberPhotos"]=>
    string(1) "6"
    ["numberLikes"]=>
    string(2) "4"
  }
}

但输出是:

array(1) {
  [0]=>
  array(2) {
    ["numberPhotos"]=>
    string(1) "2"
    ["numberLikes"]=>
    string(2) "4" (numberLikes output is right)
  }
}

那么,怎么做呢?

1 个答案:

答案 0 :(得分:1)

问题在于p.type = 2 OR p.type = 3 OR p.type = 4评估为10,因此只有2个可能的不同计数。

要解决此问题,您可以使用case声明:

COUNT(DISTINCT case when p.type in (2,3,4) then p.id end)