union all和group by怎么结合？

foxman 2004-12-21 09:43:47
select wl_wlid,kcsw_ph,wl_mc,wl_gg,wl_cz,wl_dj ,sum(kcsw_rksl)-sum(kcsw_cksl) as sqjcsl ,sum(kcsw_rkdw)-sum(kcsw_rkdw) as sqjcdw, 0 as bqrksl,0 as bqrkdw,0 as bqcksl,0 as bqrkdw,0 as bqjcsl,0 as bqjcdw
from kcsw ,wl where kcsw_wlid=wl_wlid and kcsw_jzrq <'20041219' group by wl_wlid,kcsw_ph,wl_mc,wl_gg,wl_cz,wl_dj
union all
select wl_wlid,kcsw_ph,wl_mc,wl_gg,wl_cz,wl_dj,0 as sqjcsl,0 as sqjcdw,sum(kcsw_rksl) as bqrksl,sum(kcsw_rkdw) as bqrkdw ,sum(kcsw_cksl) as bqcksl,sum(kcsw_ckdw) as bqckdw,0 as bqjcsl,0 as bqjcdw
from kcsw ,wl where kcsw_wlid=wl_wlid and kcsw_jzrq between '20041220' and '20041220'
group by wl_wlid,kcsw_ph,wl_mc,wl_gg,wl_cz,wl_dj
union all
select wl_wlid,kcsw_ph,wl_mc,wl_gg,wl_cz,wl_dj ,0 as sqjcsl,0 as sqjcdw,0 as bqrksl,0 as bqrkdw,0 as bqcksl,0 as bqckdw ,sum(kcsw_rksl)-sum(kcsw_cksl) as bqjcsl ,sum(kcsw_rkdw)-sum(kcsw_rkdw) as bqjcdw
from kcsw ,wl where kcsw_wlid=wl_wlid and kcsw_jzrq <='20041220' group by wl_wlid,kcsw_ph,wl_mc,wl_gg,wl_cz,wl_dj


如果对以上的记录再按wl_wlid,kcsw_ph,wl_mc,wl_gg,wl_cz,wl_dj 进行汇总，语句该怎么写啊？
...全文
1549 16 打赏 收藏 转发到动态 举报
AI 作业
写回复
用AI写文章
16 条回复
切换为时间正序
请发表友善的回复…
发表回复
alipaymate 2005-02-25
  • 打赏
  • 举报
 回复
up
foxman 2004-12-29
  • 打赏
  • 举报
 回复
问题解决。结帖给分
zhangzs8896 2004-12-21
  • 打赏
  • 举报
 回复
应该是没问题了吧，按你的要求，应该你最后写的可以得到正确答案了。
羊羽亻子 2004-12-21
  • 打赏
  • 举报
 回复
就是这样的了，没报错就行
foxman 2004-12-21
  • 打赏
  • 举报
 回复
谢谢各位！
这样写是否正确了？
select wl_wlid,kcsw_ph,wl_mc,wl_gg,wl_cz,wl_dj,sum(sqjcsl),sum(sqjcdw),sum(bqrksl),sum(bqrkdw),sum(bqcksl),
sum(bqckdw),sum(bqjcsl),sum(bqjcdw) from
(
select wl_wlid,kcsw_ph,wl_mc,wl_gg,wl_cz,wl_dj ,sum(kcsw_rksl)-sum(kcsw_cksl) as sqjcsl ,sum(kcsw_rkdw)-sum(kcsw_rkdw) as sqjcdw, 0 as bqrksl,0 as bqrkdw,0 as bqcksl,0 as bqckdw,0 as bqjcsl,0 as bqjcdw
from kcsw ,wl where kcsw_wlid=wl_wlid and kcsw_jzrq <'20041219' group by wl_wlid,kcsw_ph,wl_mc,wl_gg,wl_cz,wl_dj
union all
select wl_wlid,kcsw_ph,wl_mc,wl_gg,wl_cz,wl_dj,0 as sqjcsl,0 as sqjcdw,sum(kcsw_rksl) as bqrksl,sum(kcsw_rkdw) as bqrkdw ,sum(kcsw_cksl) as bqcksl,sum(kcsw_ckdw) as bqckdw,0 as bqjcsl,0 as bqjcdw
from kcsw ,wl where kcsw_wlid=wl_wlid and kcsw_jzrq between '20041220' and '20041220'
group by wl_wlid,kcsw_ph,wl_mc,wl_gg,wl_cz,wl_dj
union all
select wl_wlid,kcsw_ph,wl_mc,wl_gg,wl_cz,wl_dj ,0 as sqjcsl,0 as sqjcdw,0 as bqrksl,0 as bqrkdw,0 as bqcksl,0 as bqckdw ,sum(kcsw_rksl)-sum(kcsw_cksl) as bqjcsl ,sum(kcsw_rkdw)-sum(kcsw_rkdw) as bqjcdw
from kcsw ,wl where kcsw_wlid=wl_wlid and kcsw_jzrq <='20041220' group by wl_wlid,kcsw_ph,wl_mc,wl_gg,wl_cz,wl_dj
) a
group by wl_wlid,kcsw_ph,wl_mc,wl_gg,wl_cz,wl_dj
foxman 2004-12-21
  • 打赏
  • 举报
 回复
第一次写错了，出现了两次
正确的原文应该是：

select wl_wlid,kcsw_ph,wl_mc,wl_gg,wl_cz,wl_dj ,sum(kcsw_rksl)-sum(kcsw_cksl) as sqjcsl ,sum(kcsw_rkdw)-sum(kcsw_rkdw) as sqjcdw, 0 as bqrksl,0 as bqrkdw,0 as bqcksl,0 as bqckdw,0 as bqjcsl,0 as bqjcdw
from kcsw ,wl where kcsw_wlid=wl_wlid and kcsw_jzrq <'20041219' group by wl_wlid,kcsw_ph,wl_mc,wl_gg,wl_cz,wl_dj
union all
select wl_wlid,kcsw_ph,wl_mc,wl_gg,wl_cz,wl_dj,0 as sqjcsl,0 as sqjcdw,sum(kcsw_rksl) as bqrksl,sum(kcsw_rkdw) as bqrkdw ,sum(kcsw_cksl) as bqcksl,sum(kcsw_ckdw) as bqckdw,0 as bqjcsl,0 as bqjcdw
from kcsw ,wl where kcsw_wlid=wl_wlid and kcsw_jzrq between '20041220' and '20041220'
group by wl_wlid,kcsw_ph,wl_mc,wl_gg,wl_cz,wl_dj
union all
select wl_wlid,kcsw_ph,wl_mc,wl_gg,wl_cz,wl_dj ,0 as sqjcsl,0 as sqjcdw,0 as bqrksl,0 as bqrkdw,0 as bqcksl,0 as bqckdw ,sum(kcsw_rksl)-sum(kcsw_cksl) as bqjcsl ,sum(kcsw_rkdw)-sum(kcsw_rkdw) as bqjcdw
from kcsw ,wl where kcsw_wlid=wl_wlid and kcsw_jzrq <='20041220' group by wl_wlid,kcsw_ph,wl_mc,wl_gg,wl_cz,wl_dj
zhangzs8896 2004-12-21
  • 打赏
  • 举报
 回复
我想楼主似乎没对group by 整明白。
如楼上：
wl_wlid,kcsw_ph,wl_mc,wl_gg,wl_cz,wl_dj 进行分组，那么你select字段中除了这些字段外的其他字段必须要写在聚合函数中的。
zhangzs8896 2004-12-21
  • 打赏
  • 举报
 回复
Server: Msg 8156, Level 16, State 1, Line 1
The column 'bqrkdw' was specified multiple times for 'A'.
===
你这个字段出现多次吗？

类似：
create table tb1(a int,b varchar(2))
insert into tb1
select 10,'01' union all
select 20, '01' union all
select 30,'02'
create table tb2 (a int,b varchar(2))
insert into tb2
select 40,'01' union all
select 50, '02' union all
select 60 ,'03'

select a=sum(a),b from (
select a,b from tb1
union all
select a,b from tb2)a
group by b
drop table tb1,tb2

===
a b
----------- ----
70 01
80 02
60 03
foxman 2004-12-21
  • 打赏
  • 举报
 回复
原来的语句应该是：
select wl_wlid,kcsw_ph,wl_mc,wl_gg,wl_cz,wl_dj ,sum(kcsw_rksl)-sum(kcsw_cksl) as sqjcsl ,sum(kcsw_rkdw)-sum(kcsw_rkdw) as sqjcdw, 0 as bqrksl,0 as bqrkdw,0 as bqcksl,0 as bqckdw,0 as bqjcsl,0 as bqjcdw
from kcsw ,wl where kcsw_wlid=wl_wlid and kcsw_jzrq <'20041219' group by wl_wlid,kcsw_ph,wl_mc,wl_gg,wl_cz,wl_dj
union all
select wl_wlid,kcsw_ph,wl_mc,wl_gg,wl_cz,wl_dj,0 as sqjcsl,0 as sqjcdw,sum(kcsw_rksl) as bqrksl,sum(kcsw_rkdw) as bqrkdw ,sum(kcsw_cksl) as bqcksl,sum(kcsw_ckdw) as bqckdw,0 as bqjcsl,0 as bqjcdw
from kcsw ,wl where kcsw_wlid=wl_wlid and kcsw_jzrq between '20041220' and '20041220'
group by wl_wlid,kcsw_ph,wl_mc,wl_gg,wl_cz,wl_dj
union all
select wl_wlid,kcsw_ph,wl_mc,wl_gg,wl_cz,wl_dj ,0 as sqjcsl,0 as sqjcdw,0 as bqrksl,0 as bqrkdw,0 as bqcksl,0 as bqckdw ,sum(kcsw_rksl)-sum(kcsw_cksl) as bqjcsl ,sum(kcsw_rkdw)-sum(kcsw_rkdw) as bqjcdw
from kcsw ,wl where kcsw_wlid=wl_wlid and kcsw_jzrq <='20041220' group by wl_wlid,kcsw_ph,wl_mc,wl_gg,wl_cz,wl_dj


写成：
select wl_wlid,kcsw_ph,wl_mc,wl_gg,wl_cz,wl_dj,sqjcsl,sqjcdw,bqrksl,bqrkdw,bqcksl,bqckdw,bqjcsl,bqjcdw from
(
select wl_wlid,kcsw_ph,wl_mc,wl_gg,wl_cz,wl_dj ,sum(kcsw_rksl)-sum(kcsw_cksl) as sqjcsl ,sum(kcsw_rkdw)-sum(kcsw_rkdw) as sqjcdw, 0 as bqrksl,0 as bqrkdw,0 as bqcksl,0 as bqckdw,0 as bqjcsl,0 as bqjcdw
from kcsw ,wl where kcsw_wlid=wl_wlid and kcsw_jzrq <'20041219' group by wl_wlid,kcsw_ph,wl_mc,wl_gg,wl_cz,wl_dj
union all
select wl_wlid,kcsw_ph,wl_mc,wl_gg,wl_cz,wl_dj,0 as sqjcsl,0 as sqjcdw,sum(kcsw_rksl) as bqrksl,sum(kcsw_rkdw) as bqrkdw ,sum(kcsw_cksl) as bqcksl,sum(kcsw_ckdw) as bqckdw,0 as bqjcsl,0 as bqjcdw
from kcsw ,wl where kcsw_wlid=wl_wlid and kcsw_jzrq between '20041220' and '20041220'
group by wl_wlid,kcsw_ph,wl_mc,wl_gg,wl_cz,wl_dj
union all
select wl_wlid,kcsw_ph,wl_mc,wl_gg,wl_cz,wl_dj ,0 as sqjcsl,0 as sqjcdw,0 as bqrksl,0 as bqrkdw,0 as bqcksl,0 as bqckdw ,sum(kcsw_rksl)-sum(kcsw_cksl) as bqjcsl ,sum(kcsw_rkdw)-sum(kcsw_rkdw) as bqjcdw
from kcsw ,wl where kcsw_wlid=wl_wlid and kcsw_jzrq <='20041220' group by wl_wlid,kcsw_ph,wl_mc,wl_gg,wl_cz,wl_dj
) a
group by wl_wlid,kcsw_ph,wl_mc,wl_gg,wl_cz,wl_dj


Server: Msg 8120, Level 16, State 1, Line 1
Column 'a.sqjcsl' is invalid in the select list because it is not contained in either an aggregate function or the GROUP BY clause.
Server: Msg 8120, Level 16, State 1, Line 1
Column 'a.sqjcdw' is invalid in the select list because it is not contained in either an aggregate function or the GROUP BY clause.
Server: Msg 8120, Level 16, State 1, Line 1
Column 'a.bqrksl' is invalid in the select list because it is not contained in either an aggregate function or the GROUP BY clause.
Server: Msg 8120, Level 16, State 1, Line 1
Column 'a.bqrkdw' is invalid in the select list because it is not contained in either an aggregate function or the GROUP BY clause.
Server: Msg 8120, Level 16, State 1, Line 1
Column 'a.bqcksl' is invalid in the select list because it is not contained in either an aggregate function or the GROUP BY clause.
Server: Msg 8120, Level 16, State 1, Line 1
Column 'a.bqckdw' is invalid in the select list because it is not contained in either an aggregate function or the GROUP BY clause.
Server: Msg 8120, Level 16, State 1, Line 1
Column 'a.bqjcsl' is invalid in the select list because it is not contained in either an aggregate function or the GROUP BY clause.
Server: Msg 8120, Level 16, State 1, Line 1
Column 'a.bqjcdw' is invalid in the select list because it is not contained in either an aggregate function or the GROUP BY clause.


zhangzs8896 2004-12-21
  • 打赏
  • 举报
 回复
晕，没写全。
跟楼上基本一致
select wl_wlid,kcsw_ph,wl_mc,wl_gg,wl_cz,wl_dj
sum(..),max(..),..
from
(
你原由的sql语句
)a
group by wl_wlid,kcsw_ph,wl_mc,wl_gg,wl_cz,wl_dj
foxman 2004-12-21
  • 打赏
  • 举报
 回复
select wl_wlid ,kcsw_ph,wl_mc,wl_gg,wl_cz,wl_dj ,sqjcsl,sqjcdw,bqrksl,bqrkdw,bqcksl,bqckdw,bqjcsl,bqjcdw From
(select wl_wlid,kcsw_ph,wl_mc,wl_gg,wl_cz,wl_dj ,sum(kcsw_rksl)-sum(kcsw_cksl) as sqjcsl ,sum(kcsw_rkdw)-sum(kcsw_rkdw) as sqjcdw, 0 as bqrksl,0 as bqrkdw,0 as bqcksl,0 as bqrkdw,0 as bqjcsl,0 as bqjcdw
from kcsw ,wl where kcsw_wlid=wl_wlid and kcsw_jzrq <'20041219' group by wl_wlid,kcsw_ph,wl_mc,wl_gg,wl_cz,wl_dj
union all
select wl_wlid,kcsw_ph,wl_mc,wl_gg,wl_cz,wl_dj,0 as sqjcsl,0 as sqjcdw,sum(kcsw_rksl) as bqrksl,sum(kcsw_rkdw) as bqrkdw ,sum(kcsw_cksl) as bqcksl,sum(kcsw_ckdw) as bqckdw,0 as bqjcsl,0 as bqjcdw
from kcsw ,wl where kcsw_wlid=wl_wlid and kcsw_jzrq between '20041220' and '20041220'
group by wl_wlid,kcsw_ph,wl_mc,wl_gg,wl_cz,wl_dj
union all
select wl_wlid,kcsw_ph,wl_mc,wl_gg,wl_cz,wl_dj ,0 as sqjcsl,0 as sqjcdw,0 as bqrksl,0 as bqrkdw,0 as bqcksl,0 as bqckdw ,sum(kcsw_rksl)-sum(kcsw_cksl) as bqjcsl ,sum(kcsw_rkdw)-sum(kcsw_rkdw) as bqjcdw
from kcsw ,wl where kcsw_wlid=wl_wlid and kcsw_jzrq <='20041220' group by wl_wlid,kcsw_ph,wl_mc,wl_gg,wl_cz,wl_dj
)A
group by wl_wlid ,kcsw_ph,wl_mc,wl_gg,wl_cz,wl_dj


Server: Msg 8156, Level 16, State 1, Line 1
The column 'bqrkdw' was specified multiple times for 'A'.
long0104 2004-12-21
  • 打赏
  • 举报
 回复
select * from
(select wl_wlid,kcsw_ph,wl_mc,wl_gg,wl_cz,wl_dj ,sum(kcsw_rksl)-sum(kcsw_cksl) as sqjcsl ,sum(kcsw_rkdw)-sum(kcsw_rkdw) as sqjcdw, 0 as bqrksl,0 as bqrkdw,0 as bqcksl,0 as bqrkdw,0 as bqjcsl,0 as bqjcdw
from kcsw ,wl where kcsw_wlid=wl_wlid and kcsw_jzrq <'20041219' group by wl_wlid,kcsw_ph,wl_mc,wl_gg,wl_cz,wl_dj) a
union all
select * from
(select wl_wlid,kcsw_ph,wl_mc,wl_gg,wl_cz,wl_dj,0 as sqjcsl,0 as sqjcdw,sum(kcsw_rksl) as bqrksl,sum(kcsw_rkdw) as bqrkdw ,sum(kcsw_cksl) as bqcksl,sum(kcsw_ckdw) as bqckdw,0 as bqjcsl,0 as bqjcdw
from kcsw ,wl where kcsw_wlid=wl_wlid and kcsw_jzrq between '20041220' and '20041220'
group by wl_wlid,kcsw_ph,wl_mc,wl_gg,wl_cz,wl_dj) b
union all
select * from
(select wl_wlid,kcsw_ph,wl_mc,wl_gg,wl_cz,wl_dj ,0 as sqjcsl,0 as sqjcdw,0 as bqrksl,0 as bqrkdw,0 as bqcksl,0 as bqckdw ,sum(kcsw_rksl)-sum(kcsw_cksl) as bqjcsl ,sum(kcsw_rkdw)-sum(kcsw_rkdw) as bqjcdw
from kcsw ,wl where kcsw_wlid=wl_wlid and kcsw_jzrq <='20041220' group by wl_wlid,kcsw_ph,wl_mc,wl_gg,wl_cz,wl_dj) c
子陌红尘 2004-12-21
  • 打赏
  • 举报
 回复
select * from
(select wl_wlid,kcsw_ph,wl_mc,wl_gg,wl_cz,wl_dj ,sum(kcsw_rksl)-sum(kcsw_cksl) as sqjcsl ,sum(kcsw_rkdw)-sum(kcsw_rkdw) as sqjcdw, 0 as bqrksl,0 as bqrkdw,0 as bqcksl,0 as bqrkdw,0 as bqjcsl,0 as bqjcdw
from kcsw ,wl where kcsw_wlid=wl_wlid and kcsw_jzrq <'20041219' group by wl_wlid,kcsw_ph,wl_mc,wl_gg,wl_cz,wl_dj) a
union all
select * from
(select wl_wlid,kcsw_ph,wl_mc,wl_gg,wl_cz,wl_dj,0 as sqjcsl,0 as sqjcdw,sum(kcsw_rksl) as bqrksl,sum(kcsw_rkdw) as bqrkdw ,sum(kcsw_cksl) as bqcksl,sum(kcsw_ckdw) as bqckdw,0 as bqjcsl,0 as bqjcdw
from kcsw ,wl where kcsw_wlid=wl_wlid and kcsw_jzrq between '20041220' and '20041220'
group by wl_wlid,kcsw_ph,wl_mc,wl_gg,wl_cz,wl_dj) b
union all
select * from
(select wl_wlid,kcsw_ph,wl_mc,wl_gg,wl_cz,wl_dj ,0 as sqjcsl,0 as sqjcdw,0 as bqrksl,0 as bqrkdw,0 as bqcksl,0 as bqckdw ,sum(kcsw_rksl)-sum(kcsw_cksl) as bqjcsl ,sum(kcsw_rkdw)-sum(kcsw_rkdw) as bqjcdw
from kcsw ,wl where kcsw_wlid=wl_wlid and kcsw_jzrq <='20041220' group by wl_wlid,kcsw_ph,wl_mc,wl_gg,wl_cz,wl_dj) c
LBYYBL 2004-12-21
  • 打赏
  • 举报
 回复
select wl_wlid,kcsw_ph,wl_mc,wl_gg,wl_cz,wl_dj from
(
select wl_wlid,kcsw_ph,wl_mc,wl_gg,wl_cz,wl_dj from t1
union all
select wl_wlid,kcsw_ph,wl_mc,wl_gg,wl_cz,wl_dj from t2
)a
group by wl_wlid,kcsw_ph,wl_mc,wl_gg,wl_cz,wl_dj
didoleo 2004-12-21
  • 打赏
  • 举报
 回复
select group by wl_wlid,kcsw_ph,wl_mc,wl_gg,wl_cz,wl_dj,sum().... from

(
select wl_wlid,kcsw_ph,wl_mc,wl_gg,wl_cz,wl_dj ,sum(kcsw_rksl)-sum(kcsw_cksl) as sqjcsl ,sum(kcsw_rkdw)-sum(kcsw_rkdw) as sqjcdw, 0 as bqrksl,0 as bqrkdw,0 as bqcksl,0 as bqrkdw,0 as bqjcsl,0 as bqjcdw
from kcsw ,wl where kcsw_wlid=wl_wlid and kcsw_jzrq <'20041219' group by wl_wlid,kcsw_ph,wl_mc,wl_gg,wl_cz,wl_dj
union all
select wl_wlid,kcsw_ph,wl_mc,wl_gg,wl_cz,wl_dj,0 as sqjcsl,0 as sqjcdw,sum(kcsw_rksl) as bqrksl,sum(kcsw_rkdw) as bqrkdw ,sum(kcsw_cksl) as bqcksl,sum(kcsw_ckdw) as bqckdw,0 as bqjcsl,0 as bqjcdw
from kcsw ,wl where kcsw_wlid=wl_wlid and kcsw_jzrq between '20041220' and '20041220'
group by wl_wlid,kcsw_ph,wl_mc,wl_gg,wl_cz,wl_dj
union all
select wl_wlid,kcsw_ph,wl_mc,wl_gg,wl_cz,wl_dj ,0 as sqjcsl,0 as sqjcdw,0 as bqrksl,0 as bqrkdw,0 as bqcksl,0 as bqckdw ,sum(kcsw_rksl)-sum(kcsw_cksl) as bqjcsl ,sum(kcsw_rkdw)-sum(kcsw_rkdw) as bqjcdw
from kcsw ,wl where kcsw_wlid=wl_wlid and kcsw_jzrq <='20041220' group by wl_wlid,kcsw_ph,wl_mc,wl_gg,wl_cz,wl_dj
) a group by group by wl_wlid,kcsw_ph,wl_mc,wl_gg,wl_cz,wl_dj
zhangzs8896 2004-12-21
  • 打赏
  • 举报
 回复
select 写你的字段
from
(
)a
group by wl_wlid,kcsw_ph,wl_mc,wl_gg,wl_cz,wl_dj
union和union all的区别,union 和 union all + group by 去重效果
UNION 操作符用于合并两个或多个 SELECT 语句的结果集。 请注意，UNION 内部的 SELECT 语句必须拥有相同数量的列。列也必须拥有相似的数据类型。同时，每条 SELECT 语句中的列的顺序必须相同。 UNION 操作符选取不同的值。如果允许重复的值，请使用 UNION ALL。 sql语法: select id,`name` FROM stu union select id,`name` from stu2 在此建了两张表来比较union和union all的区别: stu
union all 和 group by 的顺序问题
通过三个实验，来看 union all 及 group by 不同的使用顺序对CPU时间及耗时的影响 1. 先Union All， 再Group By select inctmid ctmid,cnlid,inwhsid whsid,sum(qty) qty,goodsid  from (     select inctmid,cnlid,inwhsid,qty,goodsid   
hive union底层原理，union与union all再group by的性能对比
union：将多个结果集合并为一个结果集，结果集去重。 union all：将多个结果集合并为一个结果集，结果集不去重。 本次测试将四份数据去重合并，其中一份数据使用了grouping sets，所以这份数据是单独进行了一次reduce。 两种写法：一种是采用union的方法，另一种是采用union all再进行group by的方法（将不去重的结果集利用group by去重）。 接下来直...
两表union 如何保证group by 字段唯一
当要计算的指标可能来源多个表时，可能会使用到union all把不同的表中计算的指标合起来。写法一：group by之后union all ，之后再次group by保证分组字段的唯一。Union：对两个结果集进行并集操作，不包括重复行，同时进行默认规则的排序；写法二：开始不分组，将查到的分组字段union all 之后group by。Union All：对两个结果集进行并集操作，包括重复行，不进行排序；另外，回顾union和union all的区别：union会自动。
数据库-union和union all 及order by 和group by 执行顺序和group by 字段别名 及根据字段值取数据
前几天面试遇到几个面试题，分享一下 1、两个线上和线下订单表都有订单ID 和 用户，要求去重取出 所有有订单的用户，订单ID唯一 这里需要两个sql 查询的时候 都查询用户的ID 然后使用union把两个查询结果合并起来。不能使用union all，因为union all 不会去重 2、如下sql,问sql有什么问题 select ddd.id,count(1) from a dddinner join b B on ddd.id = B.id where ddd.id is not nu..
MS-SQL Server

34,838

社区成员

254,632

社区内容

发帖
与我相关
我的任务
社区描述
MS-SQL Server相关内容讨论专区
社区管理员
  • 基础类社区
  • 二月十六
  • 卖水果的net
加入社区
  • 近7日
  • 近30日
  • 至今

加载中

查看更多榜单
社区公告
暂无公告

试试用AI创作助手写篇文章吧

+ 用AI写文章