（一张表）获取没有联合的行

抱歉，我只是不知道如何简要解释我想要实现的目标。但其实很简单。

我有下表egr:

+---------+------------+
|  offid  |  groupid   |
+---------+------------+
|       1 | 101        |
|       1 | 202        |
|       2 | 202        |
|       2 | 404        |
+---------+------------+

我想获取 groupid 未链接到其他 offid 的行。结果将是：

+---------+------------+
|  offid  |  groupid   |
+---------+------------+
|       1 | 101        |
|       2 | 404        |
+---------+------------+

这可行，但我想知道是否有更优雅的方法来做到这一点？

select * from egr as egr1
where egr1.offid = 1
and egr1.groupid not in (select groupid from egr as egr2 where egr2.offid = 2 and egr1.groupid = egr2.groupid)
union
select * from egr as egr1
where egr1.offid = 2
and egr1.groupid not in (select groupid from egr as egr2 where egr2.offid = 1 and egr1.groupid = egr2.groupid)

如果您想尝试：

create table egr (offid int, groupid int);
insert into egr values (1, 101), (1, 202), (2, 202), (2, 404);

Thanks

这是你想要的吗？

select e.*
from egr e
where not exists (select 1
                  from egr e2
                  where e2.groupid = e.groupid and e2.offid <> e.offid 
                 );

或者，如果您只想限制这两个优惠：

select e.*
from egr e
where e.offid in (1, 2) and
      not exists (select 1
                  from egr e2
                  where e2.groupid = e.groupid and 
                        e2.offid in (1, 2) and
                        e2.offid <> e.offid 
                 );