admin

(一个表)获取没有联合的行

sql

对不起,我只是不知道如何简短地解释我想要达到的目标。但这实际上很简单。

我有下表egr

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

我想获取未将groupid链接到另一个offid的行。结果将是:

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

这行得通,但我想知道是否有更优雅的方法?

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);

谢谢


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2021-06-07

共1个答案

admin

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 
                 );
2021-06-07