1
我试图从一个mysql数据库中获得已经为服务付费并在某一年注册的用户数。成员可以支付多个服务,从而可以计算两次:分组通过不聚合SQL中的列组
select
count(payment_order.memberId) as members,
from_unixtime(account_login.memberFrom, '%Y') as memberFrom
from
payment_order
join
account_login
on
payment_order.memberId = account_login.memberId
Where
account_login.memberFrom != '0'
and payment_order.`status` = 'paid'
and payment_order.dateCompleted >= '2010-01-01 00:00:00'
and payment_order.dateCompleted <= '2011-12-31 23:59:59'
group by memberFrom
不是通过多年的分组,它似乎是由个人会员数来分组。我觉得我做的事情,可以;吨见树不见林:
1, 2005
4, 2005
1, 2005
1, 2006
5, 2006
5, 2006
我正在寻找的是
6, 2005
11, 2006
感激那可以解释我的befuddlement任何指针
完美的答案,解释清楚 - 会尽快它让我接受。非常感谢 – 2012-03-30 10:24:44