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?
答案 0 :(得分:1)
问题在于p.type = 2 OR p.type = 3 OR p.type = 4
评估为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
声明:
COUNT(DISTINCT case when p.type in (2,3,4) then p.id end)