左联接和计数返回意外值

Question

我有两张桌子：

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

我想要执行一个查询，返回由Igor创建的、类型为2、3或4的帖子数量以及Igor的赞数，因此，我做到了：

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'

预期产出是：

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

但产出如下：

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

那么，怎么做呢？

Answer 1

问题是，p.type = 2 OR p.type = 3 OR p.type = 4的计算结果要么是1，要么是0，因此只有两个可能的不同计数。

要解决这个问题，可以使用case语句：

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

Answer 2

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 3

这个怎么样？(对查询sql做一点修改)。

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'