那应该是你数据的问题了

7# 醉_清风

朋友要不我把数据给你，你跑跑试试

[dizzy][dizzy][dizzy][dizzy][dizzy][dizzy][dizzy][dizzy]

[em04][em04][em04][em04][em04][em04][em04][em04]

data crackman;
input date  money @;
cards;
2002.1        3
2002.1        44
2002.2        12
2002.2        12
2002.2        33
2002.3        5
2002.3        66
2002.3        5
2002.4        43
2002.4       32
2002.5       111
2002.5        23
;
run;
proc sql;
select date,money,sum(money) as summoney from crackman group by date;
quit;

老大，中间的money改去掉吧

data crackman;
input date  money @;
cards;
2002.1        3
2002.1        44
2002.2        12
2002.2        12
2002.2        33
2002.3        5
2002.3        66
2002.3        5
2002.4        43
2002.4       32
2002.5       111
2002.5        23
;
run;
/*
proc sql;
select date,money,sum(money) as summoney from crackman group by date;
quit;
*/
proc transpose data=crackman out=a;
by date;
var money;
run;
data a;
set a;
total=sum(of col1-col3);
keep date total;
run;

You can post your data, let's see what we can do for it. But i agree with the code posted

Proc sql;
create table b as
select date,sum(money) as tot_money
from a
group by date;
quit;

7# 醉_清风

程序是对的，谢谢，是我弄错了！

11# crackman


非常感谢，结果已经算出来了！

14# tinalica


yes, the code is right, I got some problems with my data, but it's solved anyway, cheers.

data a;
input date $6. money;
cards;
2002.1        3
2002.1        44
2002.2        12
2002.2        12
2002.2        33
2002.3        5
2002.3        66
2002.3        5
2002.4        43
2002.4       32
2002.5       111
2002.5        23
;run;
proc sql stimer;
create table result as
select distinct date, sum(money) as total
from a
group by date;
quit;