Question

我可以在标量函数中使用公用表表达式（CTE）吗？

我试图使用它来获取单个浮点值，但始终为null

这是我的功能代码，用于计算每位员工的总工作时间：

ALTER FUNCTION GetTotalWorkingHour 
(
    @StartDate datetime, 
    @EndDate datetime, 
    @EmpID nvarchar(6) = null
)
RETURNS float
AS
BEGIN

    DECLARE @Result float;
WITH
CTE_Start
AS
(
    SELECT  EmpID ,SUM(DATEDIFF(minute, (CAST(att.[date] AS datetime) + att.[Time]), @StartDate) *  
    CASE WHEN Funckey = 'EMPIN' THEN +1 ELSE -1 END) AS SumStart
    FROM PERS_Attendance AS att
    WHERE (EmpID = @EmpID OR @EmpID IS NULL) AND att.[date] < @StartDate GROUP BY EmpID
)
,CTE_End
AS
(
    SELECT EmpID ,SUM(DATEDIFF(minute, (CAST(att.[date] AS datetime) + att.[Time]), @EndDate) *  CASE WHEN Funckey = 'EMPIN' THEN +1 ELSE -1 END) AS SumEnd
    FROM  PERS_Attendance AS att
    WHERE (EmpID = @EmpID OR @EmpID IS NULL) AND att.[date] < @EndDate GROUP BY EmpID
)

SELECT @Result = 
    (CTE_Start.SumStart - ISNULL(CTE_End.SumEND, 0) / 60.0) --AS SumHours
FROM
        CTE_End
    LEFT JOIN CTE_Start ON CTE_Start.EmpID = CTE_End.EmpID
RETURN @Result

END
GO

上面的代码以正确的方式运行，如果我在一个查询中使用它（不在函数中），给我预期的结果，那么有什么不对？

Answer 1

我找到了解决方案，这是一个小错误，我必须添加：

SELECT @Result = 
    (SumEnd - ISNULL(SumStart, 0)) / 60.0 --AS SumHours
FROM
        CTE_End
    LEFT JOIN CTE_Start ON CTE_Start.EmpID = CTE_End.EmpID
RETURN @Result

而不是：

SELECT @Result = 
    (CTE_Start.SumStart - ISNULL(CTE_End.SumEND, 0) / 60.0) --AS SumHours
FROM
        CTE_End
    LEFT JOIN CTE_Start ON CTE_Start.EmpID = CTE_End.EmpID
RETURN @Result

标量函数中的CTE

1 个答案: