减去同一表的两行并求和

Subtract two rows of same table and sum the difference(减去同一表的两行并求和)
本文介绍了减去同一表的两行并求和的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着跟版网的小编来一起学习吧!

问题描述

OrNo    CurrentDate         PreviousDate        Finished    Amount
G988    02.05.2013 14:00:47 NULL                False       1560
G988    02.05.2013 21:30:00 02.05.2013 14:00:47 False       3170
G988    03.05.2013 06:00:00 02.05.2013 21:30:00 False       5095
G988    03.05.2013 07:46:24 03.05.2013 06:00:00 True        5254

表名:oldDate

我有这个数据,我必须计算一天的总量,但我还需要减去前一天的数量,以便只计算今天(当前日期)生产的数量.

I have this data, and I have to calculate the total amount on a single day but I need to also subtract the previous day's amount, so that only the amount which was produced today (current date) is calculated.

当前日期是处理订单的实际日期,上一个日期是处理该订单的最后一天.如果我在解释数据时不清楚,请指出我,我试过..

Current Date is the real date on which the order is processed and previous date is the last day's date on which this order was processed. Point me out if m not clear in explanation of the data, I tried ..

if t2.CurrentDate = t1.PreviousDate and datepart(t2.CurrentDate)= datepart(t1.CurrentDate) 

then
     if t1.CurrentDate>t2.CurrentDate

        then  @amount = t1.Amount

     else @amount = t2.Amount

我不擅长处理连接 .. :( 所以我对这个逻辑有问题,我尝试了其他示例中的其他代码但没有成功,任何想法都将不胜感激..

I'm bad at dealing with joins .. :( so I have problems with this logic, I tried some other code from other examples but was not successful, any ideas will be really appreciated..

推荐答案

如果你能得到一个带有列的表格:

If you could get a table with columns:

  • 或否
  • 当前日期
  • 上一个日期
  • CurrAmount
  • 上一个金额

那么解决您的问题将是微不足道的.给定 AnonymousTable,这个查询生成数据:

Then solving your problem would be trivial. Given the AnonymousTable, this query generates the data:

SELECT a.OrNo,
       a.CurrentDate AS CurrDate, a.PreviousDate AS PrevDate,
       a.Amount AS CurrAmount,    b.Amount AS PrevAmount
  FROM AnonymousTable AS a
  JOIN AnonymousTable AS b
    ON a.OrNo = b.OrNo AND a.PrevDate = b.CurrDate
UNION
SELECT a.OrNo,
       a.CurrentDate AS CurrDate, a.PreviousDate AS PrevDate,
       a.Amount AS CurrAmount,    0 AS PrevAmount
  FROM AnonymousTable AS a
 WHERE a.PreviousDate IS NULL

所以,大概,你可以写:

So, presumably, you could write:

SELECT OrNo, CurrDate, CurrAmount - PrevAmount AS NewAmount
  FROM (SELECT a.OrNo,
               a.CurrentDate AS CurrDate, a.PreviousDate AS PrevDate,
               a.Amount AS CurrAmount,    b.Amount AS PrevAmount
          FROM AnonymousTable AS a
          JOIN AnonymousTable AS b
            ON a.OrNo = b.OrNo AND a.PrevDate = b.CurrDate
        UNION
        SELECT a.OrNo,
               a.CurrentDate AS CurrDate, a.PreviousDate AS PrevDate,
               a.Amount AS CurrAmount,    0 AS PrevAmount
          FROM AnonymousTable AS a
         WHERE a.PreviousDate IS NULL
       )

同样清楚,如果你下定决心,你可以通过写作来简化事情:

Equally clearly, if you put your mind to it, you can simplify things by writing:

SELECT a.OrNo,
       a.CurrentDate AS CurrDate, a.PreviousDate AS PrevDate,
       a.Amount - b.Amount AS NewAmount
  FROM AnonymousTable AS a
  JOIN AnonymousTable AS b
    ON a.OrNo = b.OrNo AND a.PrevDate = b.CurrDate
UNION
SELECT a.OrNo,
       a.CurrentDate AS CurrDate, a.PreviousDate AS PrevDate,
       a.Amount AS NewAmount
  FROM AnonymousTable AS a
 WHERE a.PreviousDate IS NULL

这里的关键技术是 UNION 查询和自联接.您的数据具有一致的日期和时间线程化",因此当前日期和上一个日期列之间的比较是微不足道的.

The key techniques here are the UNION query and the self-join. Your data has consistent 'threading' of the dates and times, so the comparison between the current date and previous date columns is trivial.

这篇关于减去同一表的两行并求和的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持跟版网!

本站部分内容来源互联网,如果有图片或者内容侵犯您的权益请联系我们删除!

相关文档推荐

Dynamically group by with quot;Group Headerquot;(使用“组头动态分组)
How to retrieve leaf path in parent-child table in SQL Server with root ID?(如何使用根 ID 在 SQL Server 中的父子表中检索叶路径?)
Tsql union in while loop(while循环中的Tsql联合)
Pass a column as parameter to dateadd in SQL Server(将列作为参数传递给 SQL Server 中的 dateadd)
Is it possible to convert rows to a variable number of columns in T-SQL?(是否可以在 T-SQL 中将行转换为可变数量的列?)
Swap values between two rows of data(在两行数据之间交换值)