Question

您好我有一个查询，我需要显示用户每天进行的交易次数，每次交易的金额等值。

下面的查询确实这样做（通过获得平均费率找到欧元等价物），但由于货币不同，我得到的结果是货币而不是总数。查询返回的内容是：

Numb Transactions,Date, userid,transaction_type,total value (per currency),eur_equiv
1                  12/12, 2,     test              5                            10
2                   12/12,2,     test              2                             2

而我希望它返回

Numb Transactions,Date, userid,transaction_type,total value (per currency),eur_equiv
1                  12/12, 2,     test              7                           12

查询如下所示

SELECT COUNT(DISTINCT(ot.ID)) AS 'TRANSACTION COUNTER'
      ,CONVERT(VARCHAR(10) ,ot.CREATED_ON ,103) AS [DD/MM/YYYY]
      ,lad.ci
      ,ot.TRA_TYPE
    ,c.C_CODE
      ,CASE 
            WHEN op.CURRENCY_ID='CURRENCY-002' THEN SUM(CAST(op.IT_AMOUNT AS MONEY)) 
                /(
                     SELECT AVG(CAST(cr.B_RATE AS MONEY)) AS AVG_RATE
                     FROM   C_RATE cr
                     WHERE  cr.CURRENCY_ID = 'CURRENCY-002'
                 )
            WHEN op.CURRENCY_ID='-CURRENCY-005' THEN SUM(CAST(op.IT_AMOUNT AS MONEY)) 
                /(
                     SELECT AVG(CAST(cr.B_RATE AS MONEY)) AS AVG_RATE
                     FROM   C_RATE cr
                     WHERE  cr.CURRENCY_ID = 'CURRENCY-005'
                 )
            WHEN op.CURRENCY_ID='CURRENCY-006' THEN SUM(CAST(op.IT_AMOUNT AS MONEY)) 
                /(
                     SELECT AVG(CAST(cr.B_RATE AS MONEY)) AS AVG_RATE
                     FROM   C_RATE cr
                     WHERE  cr.CURRENCY_ID = 'CURRENCY-006'
                 )
                           ELSE '0'
       END AS EUR_EQUIVAL
FROM   TRANSACTION ot
       INNER JOIN PAYMENT op
            ON  op.ID = ot.ID
       INNER JOIN CURRENCY c
            ON  op.CURRENCY_ID = c.ID
       INNER JOIN ACCOUNT a
            ON  a.ID = ot.ACCOUNT_ID
       INNER JOIN ACCOUNT_DETAIL lad
            ON  lad.A_NUMBER = a.A_NUMBER
       INNER JOIN CUST cus
            ON  lad.CI = cus.CI
WHERE  ot.TRA_TYPE_ID IN ('INBANK-TYPE'
                                 ,'IN-AC-TYPE'
                                 ,'DOM-TRANS-TYPE')
       AND ot.STATUS_ID = 'COMPLETED'
       AND cus.BRANCH IN ('123'
                                      ,'456'
                                      ,'789'
                                      ,'789')
GROUP BY
       lad.CI
      ,CONVERT(VARCHAR(10) ,ot.CREATED_ON ,103)
    ,c.C_CODE
      ,op.CURRENCY_ID
      ,ot.TRAN_TYPE_ID
HAVING SUM(CAST(op.IT_AMOUNT AS MONEY))>'250000.00'
ORDER BY
       CONVERT(VARCHAR(10) ,ot.CREATED_ON ,103) ASC

Answer 1

SELECT  MIN([Numb Transactions]
        , Date
        , UserID
        , Transaction_type
        , SUM([Total Value]
        , SUM([Eur Equiv]
FROM    (
          ... -- Your current select (without order by)
        ) q
GROUP BY
        Date
        , UserId
        , Transaction_type

Answer 2

问题最常出现在联接的双行中。我所做的是先Select *，然后看看哪些列生成了双行。您可能需要调整双行的JOIN关系才能消失。

如果没有任何结果集，则很难重现您所获得的错误。检查以下事项：

Select *只返回数据上的双行，我将与聚合函数合并。如果此处的答案为“NO”，则需要使用Subselect更改JOIN关系。我在想Account 和Account detail表。
如果连接不够唯一，某些连接可以创建重复的行。也许你必须在这里加入多项内容，例如JOIN table1 ON table1.ID = table2.EXT_ID and table1.Contact = table2.Contact

Answer 3

一些事情：

1）考虑使用以下函数： SELECT AVG(CAST(cr.B_RATE AS MONEY)) AS AVG_RATE FROM C_RATE cr WHERE cr.CURRENCY_ID = 'CURRENCY-002' with the currencyID as a parameter

2）分组在这里工作吗？

3）案件或个人的总和是什么？

 sum(CASE ..) vs sum(cast(op.IT_Amount as money)

无法正确分组SQL结果

3 个答案: