LEFT JOIN 有 COUNT returns 个意外值

Question

我有两个表：

post:

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

我想查询 returns Igor 的类型 2、3 或 4 的 post 数量以及 Igor 的点赞数量，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"
  }
}

但是输出是：

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

那么，如何做到这一点？

Answer 1

尝试：

SELECT (SELECT COUNT(*) FROM post p WHERE p.type = 2 OR p.type = 3 OR p.type = 4 AND p.author = 'Igor') AS numberPhotos, (SELECT COUNT(*) FROM likes l WHERE l.auhtor = 'Igor') AS numberLikes

Answer 2

这个怎么样。（根据您的查询 sql 进行了一些修改）。

SELECT COUNT(DISTINCT p.type = 2) + COUNT(DISTINCT p.type = 3) + COUNT(DISTINCT 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'

Answer 3

问题是 p.type = 2 OR p.type = 3 OR p.type = 4 的计算结果为 1 或 0，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 语句：

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

LEFT JOIN 有 COUNT returns 个意外值

LEFT JOIN with COUNT returns unexpected value

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