Question

我的mysql数据库上有3个表（用户，关注者，帖子），我写了一个sql，从用户跟随的人那里得到所有帖子。

SQL

DaysDifference

结果

<a target="_blank" href="@baseLogsUrl/#/discover?_g=(time:(from:no@DaysDifference/d,to:now))&_a=(query:(query_string:(query:'@data._id')))">

用户表

SELECT post.* FROM post JOIN
       follower ON post.owner_user_id = follower.user_id AND
       follower.follower_id = "3"

关注者表

id | owner_user_id | content
2  | 1             | why are all my senior developers jerks?
3  | 1             | PHP7 in your face node.js

发布表格

id | username | password
 1 | user1    | 12345678
 2 | user2    | 12345678
 3 | user3    | 12345678

所以现在我正在尝试选择用户关注的帖子和用户的帖子

我试过这个sql

SQL

user_id | follower_id
3       | 1
3       | 2
1       | 3

结果

id | owner_user_id | content
1  | 2             | reading a 1k+ page on mysql, do i need to?
2  | 1             | why are all my senior developers jerks?
3  | 1             | PHP7!, in your face node.js
3  | 3             | I posted

请尝试用sql实现， if（possible）{＆＃34; what_am_i_doing_wrong＆＃34;}（）;

编辑

SELECT post.* FROM post JOIN
       follower ON post.owner_user_id = follower.user_id AND
       follower.follower_id = "3" JOIN
       user ON post.owner_user_id = user.id= "3"

Answer 1

select post.* from 
(select user_id from follower where follower_id = 3 union select 3) blah 
join post on blah.user_id = post.owner_user_id;

Answer 2

试试这个：

SELECT Distinct post.* FROM post JOIN
   follower ON post.owner_user_id = follower.user_id JOIN
   user ON follower.user_id = user.id WHERE user.id = 3

你没有在JOIN语句中使用条件，而是使用WHERE。

使用join从两个以上的表中进行选择会选择重复项

2 个答案: