给你一个oracle下的查询sql
select sell_no to_char(sell_date,'yyyymm'),
substr(to_char(last_day(sell_date),'yyyymmdd'),7,2),
sum(ell_money)
from table1
group by sell_no to_char(sell_date,'yyyymm'),
substr(to_char(last_day(sell_date),'yyyymmdd'),7,2)
select sell_no to_char(sell_date,'yyyymm'),
substr(to_char(last_day(sell_date),'yyyymmdd'),7,2),
sum(ell_money)
from table1
group by sell_no to_char(sell_date,'yyyymm'),
substr(to_char(last_day(sell_date),'yyyymmdd'),7,2)
解决方案 »
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select sell_no , to_char(sell_date,'yyyymm'),
substr(to_char(last_day(sell_date),'yyyymmdd'),7,2),
max(sell_money)
from table1
group by sell_no , to_char(sell_date,'yyyymm'),
substr(to_char(last_day(sell_date),'yyyymmdd'),7,2)
where T1.s3=select max(s3) from table1 where T1.s1=table1.s1 and T2.s2=table.s2
(select sell_no,max(sell_money)as sell_money from a group by sell_no) as b on a.sell_no=b.sell_no
left join (select sell_no,max(right(rtrim(sell_date),2)) as sell_date from a group by sell_no) as c on a.sell_no=c.sell_no方法比较苯但应可解决,数据类型问题请自行解决(可能sell_date要转换类型)
select a.sell_no,a.sell_money,b.sell_date from(
(select sell_no,max(sell_money)as sell_money, left(rtrim(sell_date),len(sell_date)-3) as mo
from a group by sell_no,left(rtrim(sell_date),len(sell_date)-3)) as a
left join(select sell_no,max(right(rtrim(sell_date),2)) as sell_date,left(rtrim(sell_date),len(sell_date)-3) as mo
from a group by sell_no,left(rtrim(sell_date),len(sell_date)-3)) as b
on a.sell_no=b.sell_no and a.mo=b.mo
)/********数据*********/
aa 2001-05-12 80
aa 2001-05-31 130
aa 2002-04-20 110
aa 2002-04-28 250
aa 2002-05-20 150
bb 2001-05-10 80
bb 2001-05-30 130
/********结果********/sell_no sell_money sell_date
---------- ----------- ---------
aa 130 31
aa 250 28
aa 150 20
bb 130 30