我知道这个问题对我所问的问题没有任何意义,但我的意思是......为了更清楚,我将用例子来说明我的问题:
我有一个userpost
表,其中包含来自不同用户的帖子。
用户发布表:
+---------+--------+--------------+
| postId | userId | postMessage |
+---------+--------+--------------+
| 1 | 3 | someText |
| 2 | 5 | someText |
| 3 | 2 | sometext |
| 5 | 6 | someText |
+---------+--------+--------------+
(!请记住,userId
被引用到users
表中(
我有另一个名为 favorites
的表,其中postId
是从userpost
表中引用的:
收藏夹表:
+---------+--------+
| postId | userId |
+---------+--------+
| 1 | 5 |
| 3 | 2 |
+---------+--------+
我想要的是从userpost
中获取所有数据,并检查某个用户帖子是否已被(WHERE userId = 5
(收藏
尝试使用此查询,但这不是我想要的!
SELECT *,
(SELECT EXISTS( SELECT * FROM `favorites` INNER JOIN `userpost` on
favourites.postId = userpost.postId WHERE favorites.postId = 1
AND favorites.userId = 5)) AS isFavourited FROM userpost;
这是以下查询的结果:
+---------+--------+-------------+--------------+
| postId | userId | postMessage | isFavourited |
+---------+--------+-------------+--------------+
| 1 | 3 | someText | 1 |
| 2 | 5 | someText | 1 |
| 3 | 2 | someText | 1 |
| 5 | 3 | someText | 1 |
+---------+--------+-------------+--------------+
我知道我通过使用以下命令在查询中犯了错误:
(WHERE favorites.postId = 1 AND favorites.userId = 5)
这确实返回 true。
我给你举个我想要的例子:
假设(userId = 5)
想要获取所有userpost
,我们必须得到以下结果:
+---------+--------+-------------+--------------+
| postId | userId | postMessage | isFavourited |
+---------+--------+-------------+--------------+
| 1 | 3 | someText | 1 |
| 2 | 5 | someText | 0 |
| 3 | 2 | someText | 0 |
| 5 | 3 | someText | 0 |
+---------+--------+-------------+--------------+
尝试:
SELECT *,
postId IN
(SELECT postId FROM favourites WHERE userId = 5)
AS isFavourited
FROM userPost
您的查询检查用户 5 收藏夹发布 1 的位置是否存在一行;而不是用户 5 收藏夹返回的该行中选择的帖子的位置。
如果我明白你在问什么,我想你可以做这样的事情:
select up.*, case when f.postId is null then "0" else "1" end as isFavourited
from userpost up
left join favourites f on f.postId = up.postId and f.userId = up.userId