Numbers of weekdays in a date range in TSQL

后端 未结 4 1973
忘了有多久
忘了有多久 2021-01-06 07:07

This is harder than it looks. I need a function that calculates the numbers of a given weekday in a date range. I don\'t want any loops or recursive SQL. There are millions

相关标签:
4条回答
  • 2021-01-06 07:29

    This is what I came up with after trying alot of different approches. I did spend a long time on solving it and I was still working on it, when I posted the question. I decided to post it as an answer because of the self-learner badge, although I never got more than 2 points for an answer.

    alter function dbo.f_countweekdays 
    ( @day int, @fromdate datetime, @todate datetime )  
    returns int 
    begin 
    RETURN (SELECT datediff(day, @fromdate, dateadd(week,datediff(week,0,@todate - 1) + 
    CASE WHEN datepart(weekday,@todate) < @day THEN 0 ELSE 1 END,0) + @day - 1) / 7)
    end
    
    0 讨论(0)
  • 2021-01-06 07:32

    An alternative approach is the good old-fashioned data warehouse time dimension, where you have a table with all potential dates in it, along with any useful information you want to filter/count by:

    Key       ActualDate  DayName   IsWeekday  DayNumberInYear  FinancialQuarter
    20110101  1 Jan 2011  Saturday  0          1                2011 Q1
    20110102  2 Jan 2011  Sunday    0          2                2011 Q1
    20110103  3 Jan 2011  Monday    1          3                2011 Q1
    

    Then just join to that table and filter, e.g.

    SELECT 
      COUNT(*) 
    FROM 
      date_dimension
    WHERE
      ActualDate BETWEEN '1 Jan 2011' AND '3 Jan 2011' AND
      IsWeekday = 1
    

    If you do date analysis a lot over a known range of dates, this can really speed up and simplify your queries. Whether you know your possible date ranges in advance is the limiting factor on whether this is helpful, really, but it's a useful trick to know about.

    0 讨论(0)
  • 2021-01-06 07:39

    @Mikael Eriksson has got a wonderful idea, but his implementation seems a bit overcomplicated.

    Here's what I've come up with (and I'd like to stress that it is based on the solution by @Mikael, to whom the main credit should go):

    ALTER FUNCTION dbo.f_countweekdays (@Dow int, @StartDate datetime, @EndDate datetime)
    RETURNS int
    AS BEGIN
      RETURN (
        SELECT
          DATEDIFF(wk, @StartDate, @EndDate)
          - CASE WHEN DATEPART(dw, @StartDate) > @Dow THEN 1 ELSE 0 END
          - CASE WHEN DATEPART(dw, @EndDate)   < @Dow THEN 1 ELSE 0 END
          + 1
      )
    END
    

    UPDATE

    As Mikael has correctly noted in his answer's comment thread, in order for the above solution to work correctly the DATEFIRST setting must be set to 7 (Sunday). Although I couldn't find this documented, a quick test revealed that DATEDIFF(wk) disregards the actual DATEFIRST setting and indeed returns the difference in weeks as if DATEFIRST was always set to 7. At the same time DATEPART(dw) does respect DATEFIRST, so with DATEFIRST set to a value other than 7 the two functions return mutually inconsistent results.

    Therefore, the above script must be amended in order to account for different values of the DATEFIRST setting when calculating DATEDIFF(wk). Happily, the fix doesn't seem to have made the solution much more complicated than before, in my opinion. Judge for yourself, though:

    ALTER FUNCTION dbo.f_countweekdays (@Dow int, @StartDate datetime, @EndDate datetime)
    RETURNS int
    AS BEGIN
      RETURN (
        SELECT
          DATEDIFF(wk, DATEADD(DAY, -@@DATEFIRST, @StartDate),
                       DATEADD(DAY, -@@DATEFIRST, @EndDate))
          - CASE WHEN DATEPART(dw, @StartDate) > @Dow THEN 1 ELSE 0 END
          - CASE WHEN DATEPART(dw, @EndDate)   < @Dow THEN 1 ELSE 0 END
          + 1
      )
    END
    

    Edited: both -@@DATEFIRST % 7 entries have been simplified to just -@@DATEFIRST, as someone suggested here.

    0 讨论(0)
  • 2021-01-06 07:43
    create function dbo.f_countweekdays
    (
      @DOW int, 
      @StartDate datetime, 
      @EndDate datetime
    ) 
    returns int
    begin
      return
      ( select datediff(wk, T2.St, T2.En) -
               case when T1.SDOW > @DOW then 1 else 0 end -
               case when T1.EDOW < @DOW then 1 else 0 end
        from (select datepart(dw, @StartDate),
                     datepart(dw, @EndDate)) as T1(SDOW, EDOW)
          cross apply (select dateadd(d, - T1.SDOW, @StartDate),
                              dateadd(d, 7 - T1.EDOW, @EndDate)) as T2(St, En))
    end
    
    0 讨论(0)
提交回复
热议问题