with t as(
select '20120411181706' time,'9999' manager,'0' type,'104' LV from dual
union all
select '20120411181706','9999','0','104' from dual
union all
select '20120411181706','9999','0','107' from dual
union all
select '20120411181706','9999','0','104' from dual
union all
select '20120411181734','9999','1','104' from dual
union all
select '20120411181734','9999','1','104' from dual
union all
select '20120411181734','9999','1','107' from dual
union all
select '20120411181734','9999','1','104' from dual
)select manager,type,lv,time,count(*) from t group by time,manager,type,lv
MANAGER TYPE LV  TIME             COUNT(*)
------- ---- --- -------------- ----------
9999    0    104 20120411181706          3
9999    0    107 20120411181706          1
9999    1    104 20120411181734          3
9999    1    107 20120411181734          1

------解决方案--------------------

探讨

我还需要根据9999找出对应的姓名
104之类的也是 要去别的表找出姓名 怎么办啊 那这个分组吗？


------解决方案--------------------

不知道这样可以不  反正效率会低很多
SQL code

select distinct manager,type,lv,time,
      (select count(1) 
        from t t1 
        where t1.manager=t.manager and t1.type=t.type and t1.lv=t.lv and t1.time=t.time) t_num
from t

------解决方案--------------------
探讨


有不用group by 用前面distinct的写法吗