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--试试这个效果一样吗 id是不是唯一的
SELECT Sum([TradeNum]), Sum([Payment]), date([TradeTime]),
sum(o.[Payment]) as NoPayment
FROM [Order]
left join Order o on o.[TradeStatus] =2 and o.ID=[Order].ID
GROUP BY date(order.tradetime)
SELECT Sum([TradeNum]), Sum([Payment]), [TradeTime],O.NoPayment
FROM [Order] O,
(select ID,sum([Payment])NoPayment from [Order] as o where [TradeStatus] =2 GROUP BY ID) T WHERE O.ID=T.ID
GROUP BY DATEPART(DD,tradetime) ,O.NoPayment
SELECT Sum([TradeNum]), Sum([Payment]), [TradeTime],O.NoPayment
FROM [Order] O,
(select ID,sum([Payment])NoPayment from [Order] as o where [TradeStatus] =2 GROUP BY ID) T WHERE O.ID=T.ID
GROUP BY date(tradetime)
--SQL 没这种写法吧
date(tradetime)