加入计算？ [英] Joins with calculation?

问题描述

select WH.date,PN.partyname,RA.recamount,(DE.totalweight*DE.rateperkg)as totalAmount,
((DE.totalweight*DE.rateperkg)-RA.recamount)as dueamount from K_RT_WarehouseDetails WH
inner join K_RT_Dailyentryretail DE on WH.sno=DE.branchdate
inner join K_RT_AmountReceived RA on RA.partyname=DE.partyname
inner join K_RT_PartyName PN on DE.partyname=PN.sno

这样输出：

Out put like this:

Date                     partyname       recamount     totalamount       dueamount
  2013-12-19 00:00:00.000     ABC             5000          15000            10000
  2013-12-20 00:00:00.000     ABC             5000          15000            14000

但我想显示像t一样的输出他的

But I want to display Output like this

Date                     partyname       recamount     totalamount       dueamount
  2013-12-19 00:00:00.000     ABC             5000          15000            10000
  2013-12-20 00:00:00.000     ABC             1000          10000             9000

如果收到金额，第二项输出总金额将会发生变化..

Second output total amount will be change ,if we recieved amount..

推荐答案

您应该使用 SUM（）聚合函数 [ ^ ]：

You should use SUM() aggregate function[^]:

SELECT WH.date,PN.partyname,SUM(RA.recamount),(DE.totalweight*DE.rateperkg)as totalAmount, (DE.totalweight*DE.rateperkg)-RA.recamount) as dueamount
FROM K_RT_WarehouseDetails WH
inner join K_RT_Dailyentryretail DE on WH.sno=DE.branchdate
inner join K_RT_AmountReceived RA on RA.partyname=DE.partyname
inner join K_RT_PartyName PN on DE.partyname=PN.sno
GROUP BY WH.date, PN.partyname
ORDER BY WH.date, PN.partyname

你的意思是你想在日期上进行分组，这样你就可以单独获得每个日期的值总数？在不知道数据是什么的情况下，你的问题毫无意义。

Do you mean you want to do group by on the date so that you get values for each date alone instead of totals ? Without knowing what the data is, you're question makes zero sense.

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