select name,sum(a) 1月份,sum(b) 2月份,sum(c) 3月份
from(
     select name,count(time) a,0 b,0 c 
     from t where to_char(time,'yyyymm') = '200701' 
     group by name
     union all
     select name,0 a,count(time) b,0 c 
     from t where to_char(time,'yyyymm') = '200702' 
     group by name
     union all
     select name,0 a,0 b,count(time) c 
     from t where to_char(time,'yyyymm') = '200703' 
     group by name
    )
group by name

------解决方案--------------------
2樓錯誤啊！


------解决方案--------------------
SQL code


create table t 
   (bm  varchar(20),
    yf  char(2),
    zc  varchar(20),
    je NUMBER);
insert into t
(select '部门A',1,'内部',10 from DUAL)
union all
(select '部门A',1,'外部',15 from DUAL)
union all
(select '部门A',2,'内部',1 from DUAL)
union all
(select '部门B',1,'内部',10 from DUAL)
union all
(select '部门B',2,'内部',10 from DUAL)
union all
(select '部门B',3,'内部',10 from DUAL);



select nvl(bm,'合计'),nvl(zc,'小计'),
  sum(decode(yf,1,je,0)) as "1月",
  sum(decode(yf,2,je,0)) as "2月",
  sum(decode(yf,3,je,0)) as "3月",
  sum(decode(yf,4,je,0)) as "4月",
  sum(decode(yf,5,je,0)) as "5月",
  sum(decode(yf,6,je,0)) as "6月",
  sum(decode(yf,7,je,0)) as "7月",
  sum(decode(yf,8,je,0)) as "8月",
  sum(decode(yf,9,je,0)) as "9月",
  sum(decode(yf,10,je,0)) as "10月",
  sum(decode(yf,11,je,0)) as "11月",
  sum(decode(yf,12,je,0)) as "12月",
  sum(je) as "合计"
from t
group by rollup(bm,zc);
  


NVL(BM,'合计')      NVL(ZC,'小计')             1月       2月       3月
---------------------------------------
部门A               内部                        10         1         0
部门A               外部                        15         0         0
部门A               小计                        25         1         0
部门B               内部                        10        10        10
部门B               小计                        10        10        10
合计                小计                        35        11        10

------解决方案--------------------
select   name,  
sum(decode(to_char(time,'MM'),'01',1,0)) Jan,  
sum(decode(to_char(time,'MM'),'02',1,0)) Feb  ,  
sum(decode(to_char(time,'MM'),'03',1,0)) Mar  
from   temp1 
group   by   name
having (sum(decode(to_char(time,'MM'),'01',1,0))
      +sum(decode(to_char(time,'MM'),'02',1,0))
      +sum(decode(to_char(time,'MM'),'03',1,0))
      )>0


------解决方案--------------------
try it ..

SQL code
 
 SQL: 
  
 select tt.na, 
    sum(decode(to_char(tt.ti,'mm'),1,1,0)) as "1 month", 
    sum(decode(to_char(tt.ti,'mm'),2,1,0)) as "2 month", 
    sum(decode(to_char(tt.ti,'mm'),3,1,0)) as "3 month",