Question

我在SQL中找到了做日期差异问题的一些答案，不包括周末和假日。我的问题是我需要进行日期比较 - 有多少子记录的工作日期是在父记录的发送日期的三天内？

大多数日期差异答案都涉及一个日历表，我想如果我可以构建一个返回日期+3的子选择，我可以解决剩下的问题。但我无法弄清楚如何返回日期+3。

所以：

CREATE TABLE calendar
(
    thedate DATETIME NOT NULL,
    isweekday SMALLINT NULL,
    isholiday SMALLINT NULL
);

和

SELECT thedate AS fromdate, xxx AS todate
FROM calendar

我想要的是todate从72小时开始，不包括周末和假日。做一个COUNT（*），其中isweekday而不是isholiday很简单，但做一个DATEADD（）是另一回事。

我不知道从哪里开始。

Answer 1

修改已更改为将非工作日包含为有效日期。

WITH rankedDates AS ( SELECT thedate , ROW_NUMBER() OVER( ORDER BY thedate ) dateRank FROM calendar c WHERE c.isweekday = 1 AND c.isholiday = 0 ) SELECT c1.fromdate , rd2.thedate todate FROM ( SELECT c.thedate fromDate , ( SELECT TOP 1 daterank FROM rankedDates rd WHERE rd.thedate <= c.thedate ORDER BY thedate DESC ) dateRank FROM calendar c ) c1 LEFT JOIN rankedDates rd2 ON c1.dateRank + 3 = rd2.dateRank

您可以在日历表上添加日期排名列以简化此操作并避免使用CTE：

CREATE TABLE calendar ( TheDate DATETIME PRIMARY KEY , isweekday BIT NOT NULL , isHoliday BIT NOT NULL DEFAULT 0 , dateRank INT NOT NULL );

然后，您只需将daterank列设置为非假日工作日。

Answer 2

这应该可以解决问题，将“顶部”中的数字更改为您想要包含的天数。

declare @date as datetime

set @date = '5/23/13'

select
    max(_businessDates.thedate)
from (
    select 
         top 3 _Calendar.thedate 
    from calendar _Calendar
    where _Calendar.isWeekday = 1 
         and _Calendar.isholiday = 0
         and _Calendar.thedate >= @date
    order by 
         _Calendar.thedate
) as _businessDates

对于可以前进或后退一定天数的动态版本，请尝试以下操作：

declare @date as datetime
declare @DayOffset as int

set @date = '5/28/13'
set @DayOffset = -3

select
    (case when @DayOffset >= 0 then 
            max(_businessDates.thedate) 
        else 
            min(_businessDates.thedate) 
        end)
from (
    select 
         top (abs(@DayOffset) + (case when @DayOffset >= 0 then 1 else 0 end)) _Calendar.thedate
    from calendar _Calendar
    where _Calendar.isWeekday = 1
        and _Calendar.isholiday = 0
        and ( (@DayOffset >= 0 and _Calendar.thedate >= @date)
        or  (@DayOffset < 0 and _Calendar.thedate < @date) )
    order by 
         cast(_Calendar.thedate as int) * (case when @DayOffset >=0 then 1 else -1 end)
) as _businessDates

您可以将@DayOffset设置为正数或负数。

Answer 3

你只需要DATEADD，除非我不理解你的问题。

DATEADD（DAY，3，没有fromdate）

编辑：我看到，不计算周末或假期，会暂时更新。

更新：看起来好像Jason钉了它，但是如果你有机会使用SQL2012，这是简单的版本：

SELECT todate = thedate
       fromdate = LEAD(thedate,3) OVER (ORDER BY thedate)
FROM calendar
WHERE isweekday = 1
 AND isHoliday = 0

Answer 4

如果您需要使用dateAdd：

作为查询，请尝试此操作

SELECT 
     allDates.thedate fromDate
    ,min(nonWeekendHoliday.thedate) toDate
FROM (    
    SELECT 
        thedate
    FROM 
       calendar _calendar
) allDates
LEFT JOIN (
    SELECT 
        thedate
    FROM 
       calendar _calendar
    WHERE 
        _calendar.isweekday = 1 
        AND 
        _calendar.isholiday = 0 
) nonWeekendHoliday
    on dateadd(d,3,allDates.thedate) <= nonWeekendHoliday.thedate
where allDates.thedate between '5/20/13' and '5/31/13'
group by
     allDates.thedate

选择日期+ 3天，不包括周末和假日

4 个答案: