Question

我们目前有以下专栏：

CustomerID         int
ArrivalDateTime    datetime
CheckPoint         varchar(20)
DepartureDateTime  datetime

CheckPoint列根据特定查询记录24小时格式时间。

CustomerID  ArrivalDateTime      CheckPoint      DepartureDateTime
1           2015-05-03 08:15     0800            2015-05-03 08:30
2           2015-05-04 13:15     1300            2015-05-04 15:30

我需要的是用CheckPoint时间替换ArrivalDateTime的时间。所以它如下所示：

CustomerID  ArrivalDateTime          DepartureDateTime
    1       2015-05-03 08:00         2015-05-03 08:30
    2       2015-05-04 13:00         2015-05-04 15:30

这样我就能从ArrivalDateTime到DepartureDateTime获得时差。预期成果：

CustomerID      ArrivalDateTime          DepartureDateTime     TimeInMinutes
        1       2015-05-03 08:00         2015-05-03 08:30      30
        2       2015-05-04 13:00         2015-05-04 15:30      150

Answer 1

试试这个：

select 
    CustomerID
    , DATEADD(day, DATEDIFF(day, 0, ArrivalDateTime), STUFF(CheckPoint_,3,0,':')) ArrivalDateTime
    , DepartureDateTime
    , DATEDIFF(mi, DATEADD(day, DATEDIFF(day, 0, ArrivalDateTime), STUFF([CheckPoint],3,0,':')), DepartureDateTime) as TimeInMinutes
from t;

如果您不想重复这些功能，可以将它们移动到cte或派生表中，如下所示：

select 
    CustomerID
    , ArrivalDateTime
    , DepartureDateTime
    , DATEDIFF(mi, ArrivalDateTime, DepartureDateTime) as TimeInMinutes
from (
    select 
        CustomerID
       , DATEADD(day, DATEDIFF(day, 0, ArrivalDateTime), STUFF([CheckPoint],3,0,':')) ArrivalDateTime
       , DepartureDateTime
    from t) a
;

Answer 2

试试这个

;WITH cte 
     AS (SELECT customerid, 
                ArrivalDateTime = Cast(CONVERT(VARCHAR(20), Cast(ArrivalDateTime AS  DATE)) 
                                       + ' ' + Stuff(CheckPoint_, 3, 0, ':') AS DATETIME), 
                departuredatetime 
         FROM   yourtable) 
SELECT customerid, 
       arrivaldatetime, 
       departuredatetime, 
       TimeInMinutes = Datediff(minute, arrivaldatetime, departuredatetime) 
FROM CTE

Answer 3

select
   CA.RevisedArrivalDateTime
 , DepartureDateTime
 , datediff(minute,CA.RevisedArrivalDateTime,DepartureDateTime)
from yourtable
cross apply (
  select dateadd(minute
                 , convert(int,left([CheckPoint],2)) * 60
                 + convert(int,right([CheckPoint],2))
                 ,dateadd(day, datediff(day,0, ArrivalDateTime ), 0))
  ) CA (RevisedArrivalDateTime)

通过将CheckPoint转换为几分钟，可以避免使用varchar转换为日期时间，从而提高效率。

此外，通过使用交叉应用，可以通过别名重用计算的新日期时间。

请参阅this sqlfiddle

Answer 4

尝试以下方法：

SELECT  CustomerID
        ,CAST(CAST(ArrivalDateTime AS DATE) AS DATETIME) + CAST(CONCAT(LEFT([CheckPoint], 2), ':', RIGHT([CheckPoint], 2)) AS DATETIME) AS ArrivalDateTime
        ,DepartureDateTime
        ,DATEDIFF(MINUTE, CAST(CAST(ArrivalDateTime AS DATE) AS DATETIME) + CAST(CONCAT(LEFT([CheckPoint], 2), ':', RIGHT([CheckPoint], 2)) AS DATETIME), DepartureDateTime) AS TimeInMinutes
FROM @tbl

带输出的示例数据

DECLARE @tbl TABLE (CustomerID INT,  ArrivalDateTime DATETIME,      [CheckPoint] Varchar(20),     DepartureDateTime DATETIME)

INSERT @tbl
SELECT 1           ,'2015-05-03 08:15',     '0800'      ,      '2015-05-03 08:30' UNION ALL
SELECT 2           ,'2015-05-04 13:15',     '1300'      ,      '2015-05-04 15:30'

SELECT  CustomerID
        ,CAST(CAST(ArrivalDateTime AS DATE) AS DATETIME) + CAST(CONCAT(LEFT([CheckPoint], 2), ':', RIGHT([CheckPoint], 2)) AS DATETIME) AS ArrivalDateTime
        ,DepartureDateTime
        ,DATEDIFF(MINUTE, CAST(CAST(ArrivalDateTime AS DATE) AS DATETIME) + CAST(CONCAT(LEFT([CheckPoint], 2), ':', RIGHT([CheckPoint], 2)) AS DATETIME), DepartureDateTime) AS TimeInMinutes
FROM @tbl

CustomerID  ArrivalDateTime         DepartureDateTime       TimeInMinutes
1           2015-05-03 08:00:00.000 2015-05-03 08:30:00.000 30
2           2015-05-04 13:00:00.000 2015-05-04 15:30:00.000 150

SQLFiddle

连接+时差

4 个答案: