Question

我正在尝试显示每个用户在本周花费时间做的事情（内部或外部工作），但时间全部在表格的同一列上，是否可以将其拆分为2个不同的列和仍然拥有它，以便它只显示每个用户一次，而不是每次他们输入的时间，这可能是整个星期多次。

下面的SQL为我提供了每个用户一周的跟踪时间，但不同行的内部和外部跟踪时间。

SELECT SUM(FilteredMag_Time.mag_hoursspent) AS Time, 
       FilteredSystemUser.fullname, 
       FilteredMag_project.mag_typename
  FROM FilteredSystemUser 
INNER JOIN FilteredMag_Task 
INNER JOIN FilteredMag_project ON FilteredMag_Task.mag_projectid = FilteredMag_project.mag_projectid 
INNER JOIN FilteredMag_Time ON FilteredMag_Task.mag_taskid = FilteredMag_Time.mag_taskid 
                            ON FilteredSystemUser.systemuserid = FilteredMag_Time.createdby
     WHERE (FilteredMag_Time.mag_starttime BETWEEN DATEADD(dd, - (DATEPART(dw, GETDATE()) - 1), GETDATE()) 
                                               AND DATEADD(dd, - (DATEPART(dw, GETDATE()) - 7), GETDATE()))
GROUP BY FilteredSystemUser.fullname, FilteredMag_project.mag_typename
ORDER BY FilteredSystemUser.fullname

以下是当前输出的示例。

Time                fullname             mag_typename
------------------ --------------------- -------------------------
1.2500000000        David Sutton        External
8.2500000000        Gayan Perera        External
9.0000000000        Paul Nieuwelaar     Internal
14.8700000000       Roshan Mehta        External
6.0000000000        Roshan Mehta        Internal
2.7800000000        Simon Phillips      External
4.6600000000        Simon Phillips      Internal

Answer 1

您可以使用SQL Server PIVOT。

像

这样的东西

DECLARE @Table TABLE(
        userID INT,
        typeID VARCHAR(20),
        TimeSpent FLOAT
)

INSERT INTO @Table SELECT 1, 'INTERNAL', 1
INSERT INTO @Table SELECT 2, 'INTERNAL', 1
INSERT INTO @Table SELECT 1, 'INTERNAL', 1
INSERT INTO @Table SELECT 1, 'INTERNAL', 1
INSERT INTO @Table SELECT 2, 'EXTERNAL', 3
INSERT INTO @Table SELECT 1, 'EXTERNAL', 3

SELECT  *
FROM 
(
    SELECT  userID, typeID, TimeSpent
    FROM    @Table
) s
PIVOT   (SUM(TimeSpent) FOR typeID IN ([INTERNAL],[EXTERNAL])) pvt

输出：

userID      INTERNAL               EXTERNAL
----------- ---------------------- ----------------------
1           3                      3
2           1                      3

Answer 2

假设FilteredMag_project.mag_typename为“内部”或“外部”，请尝试以下操作：

SELECT SUM(CASE FilteredMag_project.mag_typename 
                WHEN 'INTERNAL' THEN FilteredMag_Time.mag_hoursspent
                ELSE 0 END) AS InternalTime, 
       SUM(CASE FilteredMag_project.mag_typename 
                WHEN 'EXTERNAL' THEN FilteredMag_Time.mag_hoursspent
                ELSE 0 END) AS ExternalTime, 
       FilteredSystemUser.fullname
  FROM FilteredSystemUser 
INNER JOIN FilteredMag_Task 
INNER JOIN FilteredMag_project ON FilteredMag_Task.mag_projectid = FilteredMag_project.mag_projectid 
INNER JOIN FilteredMag_Time ON FilteredMag_Task.mag_taskid = FilteredMag_Time.mag_taskid 
                            ON FilteredSystemUser.systemuserid = FilteredMag_Time.createdby
     WHERE (FilteredMag_Time.mag_starttime BETWEEN DATEADD(dd, - (DATEPART(dw, GETDATE()) - 1), GETDATE()) 
                                               AND DATEADD(dd, - (DATEPART(dw, GETDATE()) - 7), GETDATE()))
GROUP BY FilteredSystemUser.fullname
ORDER BY FilteredSystemUser.fullname

麻烦与where子句相矛盾

2 个答案: