Question

我的过程中SQL的总和时间如何返回：

START   --    FINISH   ---   TOTAL
18:14:47    20:32:54    02:18:06
12:35:45    15:06:06    02:30:21
18:08:26    21:25:39    03:17:12

我需要总和列TOTAL（08:05:40）

我的存储过程：

Alter PROCEDURE [dbo].[timeTotal] 
@userName nvarchar(50)
AS
BEGIN

    SELECT

            CONVERT(VARCHAR(8),dateadd(HH,7,workStart),108) AS workStart,
            CONVERT(VARCHAR(8),dateadd(HH,7,workFinish),108) AS workFinish,
            CONVERT(VARCHAR(8),workFinish - workStart,108) AS total


    FROM workTime
    WHERE userName = userName


END

我需要得到08:05:40

Answer 1

要以小时计算时差，请使用以下内容：

SELECT
   (RIGHT('00' + CONVERT(VARCHAR(10), SUM(DATEDIFF(MINUTE, FromTime, ToTime)) / 60), 2)
  + ':' +
    RIGHT('00' + CONVERT(VARCHAR(2), SUM(DATEDIFF(Minute, FromTime, ToTime)) % 60), 2)
  + ':' +
    RIGHT('00' + CONVERT(VARCHAR(2), SUM(DATEDIFF(SECOND, FromTime, ToTime)) % 60), 2)) 
AS TotalTime FROM AllocateRoom

在存储过程中使用上述内容，它返回时差之和，如下所示：

01:20:20

最后您的预期数据如下：

;WITH DEMO(ID, FromTime, ToTime) AS
(
    SELECT TOP 1
        ID
       ,CONVERT(DATETIME,FromTime,120) AS FromTime
       ,CONVERT(DATETIME,ToTime,120) AS ToTime
    FROM (VALUES (1,'2016-09-03 18:14:47.000','2016-09-03 20:32:54.000')
                ,(2,'2016-09-03 12:35:45.000','2016-09-03 15:06:06.000')
    ) AS X(ID,FromTime,ToTime)
)
SELECT
    k.ID,
    k.FromTime,
    k.ToTime,
    DATEADD(SECOND,DATEDIFF(SECOND, k.FromTime, k.ToTime),CONVERT(TIME(0),'00:00:00',0)) AS TIME_DIFF,
    DATEADD(SECOND,SUM(DATEDIFF(SECOND, k.FromTime, k.ToTime)) OVER 
        (
            PARTITION BY (SELECT NULL)
        ),CONVERT(TIME(0),'00:00:00',0)) AS TIME_SUM
FROM DEMO k;

Answer 2

使用DATEDIFF（）而不是减法来查找2个日期之间的时间

DATEDIFF(SECOND, workStart, workFinish) AS total

以HH：MM：SS，

显示

CONVERT(VARCHAR(8), DATEADD(SECOND, 0, DATEDIFF(SECOND, workStart, workFinish)),108) AS total

SQL存储过程中的总和时间

2 个答案: