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SQL MS Access的嵌套子查询
EN

Stack Overflow用户
提问于 2021-06-04 20:55:39
回答 1查看 48关注 0票数 0
代码语言:javascript
复制
SELECT [doc type]
     , [customer number]
     , COUNT([customer number]) As CountCustomerNumber     
     , SUM([SumOpenAmount]) As TotalOpenAmount          
FROM 
      (SELECT d.[customer number] & d.[membership number] AS CustMemb
            , d.[customer number]
            , agg.[doc type]
            , SUM(agg.[TotalSubOpenAmount]) AS SumOpenAmount
        FROM  (SELECT [doc type]
                    , [customer number]
                    , SUM([open amount]) AS TotalSubOpenAmount
               FROM   data
               WHERE  [doc type] = 'RU' 
               GROUP BY [doc type]
                      , [customer number]
              ) agg
        INNER JOIN [data] d                                   
           ON  d.[customer number] = agg.[customer number]
        GROUP  BY d.[customer number] & d.[membership number]
                , d.[customer number]
                , agg.[doc type]
      ) AS sub
GROUP  BY [doc type]
        , [customer number]
HAVING COUNT([customer number]) = 1

我需要补充的是:

在期初金额大于0的情况下对期初金额求和,然后在期初金额<0的情况下对期初金额求和。

EN

回答 1

Stack Overflow用户

发布于 2021-06-04 21:28:09

尝试条件求和:

代码语言:javascript
复制
SELECT [doc type]
     , [customer number]
     , COUNT([customer number]) As CountCustomerNumber
     , SUM(IIF([open_amount]>0, [open_amount], 0)) AS sum_open_amount_pos   
     , SUM(IIF([open_amount]<0, [open_amount], 0)) As sum_open_amount_neg         
FROM 
      (SELECT d.[customer number] & d.[membership number] AS CustMemb
            , d.[customer number]
            , agg.[doc type]
            , SUM(agg.[TotalSubOpenAmount]) AS SumOpenAmount
        FROM  (SELECT [doc type]
                    , [customer number]
                    , SUM([open amount]) AS TotalSubOpenAmount
               FROM   data
               WHERE  [doc type] = 'RU' 
               GROUP BY [doc type]
                      , [customer number]
              ) agg
        INNER JOIN [data] d                                   
           ON  d.[customer number] = agg.[customer number]
        GROUP  BY d.[customer number] & d.[membership number]
                , d.[customer number]
                , agg.[doc type]
      ) AS sub
GROUP  BY [doc type]
        , [customer number]
HAVING COUNT([customer number]) = 1

您需要首先获取未结金额字段。

代码语言:javascript
复制
SELECT [doc type]
     , [customer number]
     , COUNT([customer number]) As CountCustomerNumber
     , SUM(IIF([open_amount]>0, [open_amount], 0)) AS sum_open_amount_pos   
     , SUM(IIF([open_amount]<0, [open_amount], 0)) As sum_open_amount_neg         
FROM 
      (
-- This is the part in which the field open_amount should be fetched.
) AS sub
GROUP  BY [doc type]
        , [customer number]
HAVING COUNT([customer number]) = 1
票数 0
EN
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接:

https://stackoverflow.com/questions/67837628

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