根据两列计算求和字段

Calculate a sum field based on two columns

我有一个访问查询正在尝试 运行,但似乎无法正确处理。

我已经在计算按名称和描述列分组的总小时数。我希望总计列基本上按名称和描述列

给出总小时数
SELECT dbo_t_SAP_AttCodes.Description, Sum(dbo_v_MES_TcActivities.Hours) AS SumOfHours, 
    Left([supervisor1email],InStr([supervisor1email],".")-1) AS [Supervisor First Name], 
    Mid([supervisor1email],InStr([supervisor1email],".")+1,
    InStr([supervisor1email],"@")-InStr([supervisor1email],".")-1) 
    AS [Supervisor Last Name], (SELECT sum([hours]) FROM [dbo_v_MES_TcActivities]     
    WHERE [costctr]="106330" AND [clockin] Between DateAdd("d",-((Weekday(Date())-1)),
    Date()) And Date()) AS Total_Hours

FROM dbo_v_MES_TcActivities LEFT JOIN dbo_t_SAP_AttCodes 
    ON dbo_v_MES_TcActivities.AttCode = dbo_t_SAP_AttCodes.Code

WHERE (((dbo_v_MES_TcActivities.AttCode) Not Like "MEAL") 
    AND ((dbo_v_MES_TcActivities.CostCtr) Like "106330") 
    AND ((dbo_v_MES_TcActivities.ClockIn) Between DateAdd("d",-((Weekday(Date())-1)),
    Date()) And Date()))

GROUP BY dbo_t_SAP_AttCodes.Description, Left([supervisor1email],
    InStr([supervisor1email],".")-1), 
    Mid([supervisor1email],InStr([supervisor1email],".")+1,
    InStr([supervisor1email],"@")-InStr([supervisor1email],".")-1), 
    dbo_v_MES_TcActivities.Supervisor1Email

ORDER BY Mid([supervisor1email],InStr([supervisor1email],".")+1,
    InStr([supervisor1email],"@")-InStr([supervisor1email],".")-1);

编辑: 这是当前输出的样子:

before

以下是我希望看到的内容: After

考虑将您的子查询转换为相关子查询,在该子查询中,您将嵌套的 SELECT 语句与主查询中相应行的名称相匹配,而无需 Description(仅 姓名)分组。由于 Supervisor 的名字和姓氏计算表达式,下面的内容很长。

请注意:下面使用 table 别名 submain(相应地调整所有字段引用):

子查询

SELECT --...same fields/expressions as original w/ 'main' alias...,
      (SELECT SUM(sub.[hours]) FROM [dbo_v_MES_TcActivities] sub    
       WHERE sub.[costctr]="106330" 
         AND sub.AttCode Not Like "MEAL"
         AND sub.[clockin] Between DateAdd("d",-((Weekday(Date())-1)), Date()) And Date()
         AND Left(sub.[supervisor1email], InStr(sub.[supervisor1email],".")-1) = 
             Left(main.[supervisor1email], InStr(main.[supervisor1email],".")-1) 

         AND Mid(sub.[supervisor1email], InStr(sub.[supervisor1email],".")+1,
                 InStr(sub.[supervisor1email],"@") - InStr(sub.[supervisor1email],".")-1) =
             Mid(main.[supervisor1email], InStr(main.[supervisor1email],".")+1, 
                 InStr(main.[supervisor1email],"@") - InStr(main.[supervisor1email],".")-1)
       ) AS Total_Hours
FROM dbo_v_MES_TcActivities main
LEFT JOIN dbo_t_SAP_AttCodes sap
      ON main.AttCode = sap.Code
--...same WHERE and GROUPBY as original w/ 'main' and 'sap' aliases...

派生表

或者,将两个聚合查询都转换为派生的 table 并连接到主查询中。在 MS Access 中,您甚至可以将两个嵌套的 SELECT 语句另存为单独的已保存查询,并在 FROMJOIN 子句中引用它们。再次注意 table 别名:

SELECT t1.Description, t1.SumOfHours, t1.[Supervisor First Name], t1.[Supervisor Last Name], 
       t2.TotalHours    
FROM (
      SELECT --...same fields/expressions as original...
      FROM dbo_v_MES_TcActivities main
      LEFT JOIN dbo_t_SAP_AttCodes sap
           ON main.AttCode = sap.Code
      --...same WHERE and GROUPBY as original...
      ) As t1
INNER JOIN
      (
       SELECT SUM(sub.[hours]),
              Left(sub.[supervisor1email], InStr(sub.[supervisor1email],".")-1) 
                    AS [Supervisor First Name],
              Mid(sub.[supervisor1email], InStr(sub.[supervisor1email],".")+1,
                 InStr(sub.[supervisor1email],"@") - InStr(sub.[supervisor1email],".")-1) 
                    AS [Supervisor Last Name]
       FROM [dbo_v_MES_TcActivities] sub    
       WHERE sub.[costctr]="106330" 
         AND sub.AttCode Not Like "MEAL"
         AND sub.[clockin] Between DateAdd("d",-((Weekday(Date())-1)), Date()) And Date()
       GROUP BY
              Left(sub.[supervisor1email], InStr(sub.[supervisor1email],".")-1),
              Mid(sub.[supervisor1email], InStr(sub.[supervisor1email],".")+1,
                  InStr(sub.[supervisor1email],"@") - InStr(sub.[supervisor1email],".")-1)
       ) AS t2
ON t1.[Supervisor First Name] = t2.[Supervisor First Name]
AND t1.[Supervisor Last Name] = t2.[Supervisor Last Name]