Question

我有两张桌子：

[TABLE1]

+----------+-------------------+
| USERNAME | POST              |
+----------+-------------------+
| Bob      | 'Hi There'        |
| Jack     | 'Hello'           |
| Bob      | 'Today is Monday' |
+----------+-------------------+


[TABLE2]

+----------+-----------+
| USERNAME | FOLLOWING |
+----------+-----------+
| Mike     | Jack      |
| Jack     | Bob       |
| Bob      | Jack      |
| Jack     | Mike      |
+----------+-----------+

我知道当你想要一行时如何加入两个表..但是当你想要解决下面的问题时呢？

Select * from TABLE1 when TABLE1.USERNAME IS A USERNAME JACK is FOLLOWING;

Answer 1

对于用户名杰克如下：

select TABLE1.USERNAME, TABLE1.POST 
from TABLE1, TABLE2
where 
     TABLE1.USERNAME = TABLE2.FOLLOWING 
     and
     TABLE2.USERNAME = "Jack"

对于关注杰克的用户名

select TABLE1.USERNAME, TABLE1.POST 
from TABLE1, TABLE2
where 
     TABLE1.USERNAME = TABLE2.USERNAME 
     and
     TABLE2.FOLLOWING = "JACK"

Answer 2

尝试（假设PHP）

$username - 'Jack';
"SELECT t1.* FROM TABLE1 t1
INNER JOIN table2 t2 
ON t1.username = t2.username AND t2.following = '".$username."' ";

注意：两个表格的列名相同，您可以使用USING(username)代替ON条件

Answer 3

这有帮助吗？

SELECT * FROM table1 t1, table2 t2
WHERE t2.username = 'Jack'
AND t1.username = t2.following

跟随用户时获取mysql记录

3 个答案: