[英]fetch all the unique record from table
I have an mysql query. 我有一个mysql查询。 I need fetch the all "eid" where "aid" not equal to "15" .
我需要获取所有的“ eid” ,其中“ aid”不等于“ 15” 。 My table is like below.
我的桌子如下。 The result should be in group.
结果应在组中。 Example if aid != 15 then We should get only "eid" = 3 and 4 .
例如,如果aid!= 15,那么我们应该只获得“ eid” = 3和4 。
id | eti | aid | eid | val
1 | 2 | 15 | 1 | WDC
2 | 2 | 11 | 1 | USA
3 | 2 | 9 | 1 | XYX
4 | 2 | 15 | 2 | LDN
5 | 2 | 11 | 2 | UK
6 | 2 | 9 | 2 | ABC
7 | 2 | 11 | 3 | China
8 | 2 | 9 | 3 | HNB
9 | 2 | 18 | 4 | China
10 | 2 | 12 | 4 | HNB
Anybody help me? 有人帮我吗 Thanks in advance...
提前致谢...
select eid
from your_table
group by eid
having sum(aid = 15) = 0
The query below should help you: 以下查询将为您提供帮助:
select 'aid' from 'table'
where 'eid' not in
(select 'eid' from 'table'
where 'aid'='15')
group by 'eid';
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