SQL函数根据持续时间不工作计算结束时间

时间:2016-09-02 19:17:55

标签: sql sql-server function datetime ssms

您能否帮助我解决有关在T-SQL中编写标量函数脚本的问题?我有来自呼叫中心的一些数据,其中包含每个呼叫的开始时间和持续时间。我想创建一个输出调用结束时间的函数。以下是数据样本:

CALLID     EMPLOYEE_ID   START_TIME               DURATION     STATUS      
--------  ----------    -----------------------  ------------ ------------
796544     205285       2016-07-29 19:29:02.000  00:00:27     Complete    
796543     205284       2016-07-29 19:25:31.000  00:02:31     Complete    
796542     205284       2016-07-29 19:22:01.000  00:00:50     Complete    
796541     205285       2016-07-29 19:11:58.000  00:00:21     Complete    
796540     205285       2016-07-29 19:07:40.000  00:02:16     Complete 

我已经创建了一个我认为应该可以工作的函数,但它总是返回开始时间而不添加任何内容。我希望它解析[DURATION](varchar),以秒为单位计算持续时间,然后将其添加到开始时间。这是它的样子:

ALTER FUNCTION [dbo].[AddDuration] (@Duration varchar(8), @StartTime datetime)
RETURNS datetime
BEGIN

DECLARE @Hours      int     = CONVERT(int,SUBSTRING(@Duration,1,2))
DECLARE @Minutes    int     = CONVERT(int,SUBSTRING(@Duration,4,2))
DECLARE @Seconds    int     = CONVERT(int,SUBSTRING(@Duration,7,2))
DECLARE @EndTime    datetime

SET @EndTime = DATEADD(SECOND,(@Hours * 60 * 60) + (@Minutes * 60) + @Seconds, @StartTime)

RETURN (@EndTime)
END

我错过了什么?当我在SELECT语句中执行完全相同的逻辑时,它完美地运行。但功能并没有。

SELECT  [START_TIME]
    ,   [END_SELECT] = (DATEADD(SECOND,
                   (CONVERT(int,SUBSTRING(DURATION,1,2))*60*60)
                 + (CONVERT(int,SUBSTRING(DURATION,4,2))*60)
                 + (CONVERT(int,SUBSTRING(DURATION,7,2)))
                ,  START_TIME)
    )
    ,   [END_FUNCTION] = dbo.AddDuration(DURATION,START_TIME)


START_TIME                END_SELECT                END_FUNCTION
-----------------------   -----------------------   -----------------------
2016-08-25 09:21:00.000   2016-08-25 09:24:55.000   2016-08-25 09:21:00.000
2016-08-25 09:26:00.000   2016-08-25 09:31:22.000   2016-08-25 09:26:00.000

感谢您的帮助!

2 个答案:

答案 0 :(得分:4)

您可以一起添加两个日期时间。并且,您的时间看起来像是一个很好的适当时间格式。你可能会考虑:

select start_time + cast(cast(duration as time) as datetime)
from t;

这似乎比udf更容易。

答案 1 :(得分:0)

试试这个:

CREATE TABLE #TABLE1 (CALLID INT,EMPLOYEE_ID   INT,START_TIME    DATETIME,DURATION   TIME,  STATUS      VARCHAR(15))

INSERT INTO #TABLE1

SELECT 796544, 205285, '2016-07-29 19:29:02.000',  '00:00:27', 'Complete' UNION ALL
SELECT 796543, 205284, '2016-07-29 19:25:31.000',  '00:02:31', 'Complete' UNION ALL    
SELECT 796542, 205284, '2016-07-29 19:22:01.000',  '00:00:50', 'Complete' UNION ALL    
SELECT 796541, 205285, '2016-07-29 19:11:58.000',  '00:00:21', 'Complete' UNION ALL    
SELECT 796540, 205285, '2016-07-29 19:07:40.000',  '00:02:16', 'Complete' 

SELECT *,DATEADD(SECOND, 
              DATEPART(HOUR, DURATION)*60*60 -- HOURS TO SECONDS
            + DATEPART(MINUTE, DURATION)*60  -- MINUTES TO SECONDS
            + DATEPART(SECOND,DURATION)      -- SECONDS
            , START_TIME) END_TIME
FROM #TABLE1