计算在SQL Server中排除周末(周一到周五)的天数

怎样计较SQL Server 2008中表(从第1行到末了)的两个日期之间的事变天数？

我试过这样的对象,但它不起浸染

DECLARE @StartDate as DATETIME,@EndDate as DATETIME

Select @StartDate = date2 from testtable ;
select @EndDate = date1 from testtable ;

SELECT
   (DATEDIFF(dd,@StartDate,@EndDate) + 1)
  -(DATEDIFF(wk,@EndDate) * 2)
  -(CASE WHEN DATENAME(dw,@StartDate) = 'Sunday' THEN 1 ELSE 0 END)
  -(CASE WHEN DATENAME(dw,@EndDate) = 'Saturday' THEN 1 ELSE 0 END)

办理要领

SELECT  COUNT(*)
FROM    dbo.CalendarTable
WHERE   IsWorkingDay = 1
AND     [Date] > @StartDate
AND     [Date] <= @EndDate;

因为SQL不相识国定沐日,譬喻两个日期之间的事变日数并不老是代表事变日数.这就是大大都数据库必需行使日历表的缘故起因.它们不占用大量内存并简化了大量查询.

但假如这不是一个选项,那么你可以在运行中相对轻易地天生一个日期表并行使它

SET DATEFIRST 1;
DECLARE @StartDate DATETIME = '20131103',@EndDate DATETIME = '20131104';

-- GENERATE A LIST OF ALL DATES BETWEEN THE START DATE AND THE END DATE
WITH AllDates AS
(   SELECT  TOP (DATEDIFF(DAY,@EndDate))
            D = DATEADD(DAY,ROW_NUMBER() OVER(ORDER BY a.Object_ID),@StartDate)
    FROM    sys.all_objects a
            CROSS JOIN sys.all_objects b
)
SELECT  WeekDays = COUNT(*)
FROM    AllDates
WHERE   DATEPART(WEEKDAY,D) NOT IN (6,7);

编辑

假如您必要计较两个日期列之间的差别,您如故可以行使您的日历表：

SELECT  t.ID,t.Date1,t.Date2,WorkingDays = COUNT(c.DateKey)
FROM    TestTable t
        LEFT JOIN dbo.Calendar c
            ON c.DateKey >= t.Date1
            AND c.DateKey < t.Date2
            AND c.IsWorkingDay = 1
GROUP BY t.ID,t.Date2;

Example on SQL-Fiddle

（编辑：湖南网）

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