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作为复杂查询的一部分，我正在尝试对 DifferentUsers 字段求和：

在此处输入图像描述

这是实际的表，我使用的查询是：

SELECT '3' AS RowType
    ,DTH.PointPerson AS Person
    ,COALESCE(PDT.[Name], APP.AppName) AS Project
    ,(
        CASE WHEN (
                    STY.KanBanProductId IS NOT NULL
                    AND STY.SprintId IS NULL
                    ) THEN 'KanBan' WHEN (
                    STY.KanBanProductId IS NULL
                    AND STY.SprintId IS NOT NULL
                    ) THEN 'Sprint' ELSE SCY.Catagory END
        ) AS ProjectType
    ,COALESCE(STY.[Number], NSS.IncidentNumber) AS StoryNumber
    ,COALESCE(STY.Title, NSS.[Description]) AS StoryTitle
    ,CONVERT(VARCHAR(20), STY.Effort) AS Effort
    ,COALESCE(TSK.[Name], '') AS Task
    ,CONVERT(VARCHAR(20), TSK.OriginalEstimateHours) AS OriginalEstimateHours
    ,SCY.Catagory AS Category
    ,NSS.IncidentNumber AS IncidentNumber
    ,APP.AppName AS ApplicationName
    ,CAST(SUM(CASE WHEN DATEPART(dw, DTH.ActivityDate) = 2 THEN DTH.[Hours] ELSE 0 END) AS VARCHAR(20)) AS MondayHours
    ,CAST(SUM(CASE WHEN DATEPART(dw, DTH.ActivityDate) = 3 THEN DTH.[Hours] ELSE 0 END) AS VARCHAR(20)) AS TuesdayHours
    ,CAST(SUM(CASE WHEN DATEPART(dw, DTH.ActivityDate) = 4 THEN DTH.[Hours] ELSE 0 END) AS VARCHAR(20)) AS WednesdayHours
    ,CAST(SUM(CASE WHEN DATEPART(dw, DTH.ActivityDate) = 5 THEN DTH.[Hours] ELSE 0 END) AS VARCHAR(20)) AS ThursdayHours
    ,CAST(SUM(CASE WHEN DATEPART(dw, DTH.ActivityDate) = 6 THEN DTH.[Hours] ELSE 0 END) AS VARCHAR(20)) AS FridayHours
    ,CAST(SUM(CASE WHEN DATEPART(dw, DTH.ActivityDate) = 7 THEN DTH.[Hours] ELSE 0 END) AS VARCHAR(20)) AS SaturdayHours
    ,CAST(SUM(CASE WHEN DATEPART(dw, DTH.ActivityDate) = 1 THEN DTH.[Hours] ELSE 0 END) AS VARCHAR(20)) AS SundayHours
    ,CAST(SUM(DTH.[Hours]) AS VARCHAR(20)) AS TotalHours
    ,CAST(SUM(DTH.[UserDifference]) AS VARCHAR(20)) AS DifferentUsers
FROM DailyTaskHours DTH
LEFT JOIN Task TSK ON DTH.TaskId = TSK.PK_Task
LEFT JOIN Story STY ON TSK.StoryId = STY.PK_Story
LEFT JOIN NonScrumStory NSS ON DTH.NonScrumStoryId = NSS.PK_NonScrumStory
LEFT JOIN SupportCatagory SCY ON NSS.CatagoryId = SCY.PK_SupportCatagory
LEFT JOIN [Application] APP ON NSS.ApplicationId = APP.PK_Application
LEFT JOIN Sprint SPT ON STY.SprintId = SPT.PK_Sprint
LEFT JOIN Product PDT ON STY.ProductId = PDT.PK_Product
LEFT JOIN [User] USR ON DTH.PointPerson = USR.DisplayName
WHERE DTH.PointPerson LIKE @userParam
    AND ActivityDate >= @startDateParam
    AND ActivityDate <= @endDateParam
    AND PDT.PK_Product LIKE @productId
    AND (
        (
            @orgTeamPK = '%'
            AND (
                USR.[OrganizationalTeamId] LIKE @orgTeamPK
                OR USR.[OrganizationalTeamId] IS NULL
                )
            )
        OR (
            @orgTeamPK <> '%'
            AND (USR.[OrganizationalTeamId] LIKE @orgTeamPK)
            )
        AND (
            (
                STY.Number LIKE @search
                OR STY.Number IS NULL
                )
            OR (
                STY.Title LIKE @search
                OR STY.Number IS NULL
                )
            OR (
                TSK.NAME LIKE @search
                OR STY.Number IS NULL
                )
            )
        )
GROUP BY DTH.PointPerson
    ,PDT.[Name]
    ,SPT.[Name]
    ,SPT.[Description]
    ,STY.[Number]
    ,STY.Title
    ,TSK.[Name]
    ,SCY.Catagory
    ,NSS.IncidentNumber
    ,APP.AppName
    ,STY.KanBanProductId
    ,STY.SprintId
    ,NSS.[Description]
    ,TSK.OriginalEstimateHours
    ,STY.Effort
HAVING SUM(DTH.[Hours]) > 0

我要更改的查询部分是：

,CAST(SUM(DTH.[UserDifference]) AS VARCHAR(20)) AS DifferentUsers

如您所见，当前查询按预期执行：

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但是，如果 Hours > 0，我只希望 UserDifference 求和。

我试过这样的事情：

 ,CAST(SUM(DTH.[UserDifference]) AS VARCHAR(20) WHERE DTH.Hours > 0) AS DifferentUsers

但是语法关闭了，我收到了错误。如何实现所需的功能？

4

1 回答 1

1

你想要一个条件SUM()并且可以使用一个CASE语句：

,CAST(SUM(CASE WHEN DTH.Hours > 0 THEN DTH.[UserDifference] END) AS VARCHAR(20)) AS DifferentUsers

注意：有些人更喜欢设置ELSE，但默认情况下NULL，任何条件都没有捕获值CASE，并且由于聚合排除NULL了您不需要的值，因此上面和下面将返回相同的结果：

,CAST(SUM(CASE WHEN DTH.Hours > 0 THEN DTH.[UserDifference] ELSE 0 END) AS VARCHAR(20)) AS DifferentUsers
于 2013-11-05T15:23:07.760 回答