从一个mysql表中选择，其中id匹配来自另一个mysql表的select的结果

Question

我知道我可以在PHP中实现这一点，但我真的需要在单个查询中使用它以便更轻松地进行分页。我有两个表如下：

S_MATTERS

id | name
---------------------
 0 | Client 1
 1 | Client 2
 2 | Client 3

S_LINKS

mid | uid
---------------------
 2  | 0007
 0  | 0007

所以我想

select * from s_matters where id = (select mid from s_links where uid = 0007)

显然这不是正确的语法，我只需要从s_matters表中获取所有客户端名称，其中uid是另一个表上的0007（s_matters = mid上s_links上的id）。

Answer 1

select m.*
    from s_matters m
        inner join s_links l
            on m.id = l.mid
    where l.uid = '0007'

Answer 2

尝试将“=”更改为“in”

 select * from s_matters where id in (select mid from s_links where uid = 0007)

Answer 3

SELECT * FROM s_matters JOIN s_links ON ( s_matters.id = s_links.mid ) WHERE s_links.uid = '0007'

Answer 4

您正在寻找内部联接

select * from s_matters 
  inner join s_links on s_links.mid = s_matters.id 
where s_links.uid = '0007'