Question

我有两张桌子，sfuser和会员资格

sfuser得到139，会员桌得到50

我需要查询来挑选谁不是任何项目的成员，这意味着89名成员

这是我的查询

select sfuser.id,sfuser.username,sfuser.email from sfuser INNER JOIN projectmembership ON sfuser.id = projectmembership.member_id ;

但是我得到了50名成员我想要反过来

请小心点亮

最诚挚的问候 Sathish所在

Answer 1

select 
sfuser.id,
sfuser.username,
sfuser.email 
from sfuser 
LEFT JOIN projectmembership ON sfuser.id = projectmembership.member_id
WHERE 
projectmembership.member_id is NULL

它应该返回项目成员表中没有的所有用户。

Answer 2

INNER JOIN将只生成与两个表匹配的行。您可以执行左连接并过滤掉空白记录。类似的东西：

SELECT sfuser.id,
sfuser.username,
sfuser.email
FROM sfuser LEFT JOIN projectmembership ON sfuser.id = projectmembership.member_id
WHERE (projectmembership.member_id is null);

Answer 3

内部联接将返回在两个表之间共享完成连接的值的所有行。（见this chart）

您要查找的是左连接，即左表中右侧没有匹配的所有行。

select sfuser.id,
  sfuser.username,
  sfuser.email 
from sfuser 
LEFT JOIN projectmembership 
  ON sfuser.id = projectmembership.member_id

Answer 4

如果sfuser记录在项目成员资格表中可能有多次出现，那么就忘记外连接 - NOT EXISTS就是你想要的：

select sid,
       username,
       email
from   sfuser
where  not exists (
         select null
         from   projectmembership
         where  sfuser.id = projectmembership.member_id);

SQL连接查询左表中不匹配的行

4 个答案: