name,charge,date姓名,费用,日期
username1,0,2012-04-01
username1,10,2012-04-01
username1,0,2012-04-01username2,0,2012-04-01
username2,0,2012-04-01
username2,20,2012-04-01想把今天的给取出来:
结果如下:username1,10
username2,20
username1,0,2012-04-01
username1,10,2012-04-01
username1,0,2012-04-01username2,0,2012-04-01
username2,0,2012-04-01
username2,20,2012-04-01想把今天的给取出来:
结果如下:username1,10
username2,20
username1,0,2012-04-01
username1,10,2012-04-01
username1,0,2012-04-01
username1,0,2012-04-01
username2,0,2012-04-01
username2,0,2012-04-01
username2,20,2012-04-01
username2,0,2012-04-01就找这种出现过四条的记录才这么做.
想把今天的给取出来:
结果如下:username1,10
username2,20
select 姓名,sum(费用),日期 from tt group by 姓名,日期 having count(*)>=4 and sum(费用)>0