select grade,depname,zyname,pubnum,pbname,pbprice,pbdiscount,sum(enum)enum,sum(eprice)eprice from 表 group by grade,depname,zyname,pubnum,pbname,pbprice,pbdiscount
select grade,depname,zyname,pubnum,pbname,pbprice,pbdiscount,sum(Enum) as enum,sum(Eprice) as eprice from TB group by grade,depname,zyname,pubnum,pbname,pbprice,pbdiscount简单点儿
试试这个: select grade,depname,zyname,pubnum,pbname,pbprice,pbdiscount, SUM(enum) as enum, SUM(eprice) as eprice from tb group by grade,depname,zyname,pubnum,pbname,pbprice,pbdiscount
原来是用Group by,谢谢几位了!
恭迎品鉴select a.*,b.enum,b.eprice from ( select distinct grade,depname,zyname,pubnum,pbname,pbprice,pbdiscount from 表 ) a inner join ( select distinct pubnum,sum(enum) enum,sum(eprice) eprice from 表 group by pubnum ) b on a.pubnum=b.pubnum
select a.*,b.enum,b.eprice from ( select distinct grade,depname,zyname,pubnum,pbname,pbprice,pbdiscount from 表 ) a inner join ( select pubnum,sum(enum) enum,sum(eprice) eprice from 表 group by pubnum ) b on a.pubnum=b.pubnum
from 表
group by grade,depname,zyname,pubnum,pbname,pbprice,pbdiscount
from TB
group by grade,depname,zyname,pubnum,pbname,pbprice,pbdiscount简单点儿
depname ,
zyname ,
pubnum ,
pbname ,
pbprice ,
pbdiscount ,
SUM(enum) enum ,
SUM(eprice) eprice
FROM 表
GROUP BY grade ,
depname ,
zyname ,
pubnum ,
pbname ,
pbprice ,
pbdiscount
select grade,depname,zyname,pubnum,pbname,pbprice,pbdiscount,
SUM(enum) as enum,
SUM(eprice) as eprice
from tb
group by grade,depname,zyname,pubnum,pbname,pbprice,pbdiscount
原来是用Group by,谢谢几位了!
(
select distinct grade,depname,zyname,pubnum,pbname,pbprice,pbdiscount
from 表
) a
inner join
(
select distinct pubnum,sum(enum) enum,sum(eprice) eprice
from 表
group by pubnum
) b
on a.pubnum=b.pubnum
select a.*,b.enum,b.eprice from
(
select distinct grade,depname,zyname,pubnum,pbname,pbprice,pbdiscount
from 表
) a
inner join
(
select pubnum,sum(enum) enum,sum(eprice) eprice
from 表
group by pubnum
) b
on a.pubnum=b.pubnum