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Generate Calendar Dates. Version 2.0

Concerning my howto on sqljunkies I've got a feedback from Thomas Darmstandler. Thanks, Tom. He points out some things I didn't make up to the end. These are:


1) not returning 365 or 366 days for the year
2) although my version #1 takes care of a leap year, 
you still need to figure out whether to return 365 or 366 days

The following function by Tom returns the days of the year you requested.


alter function fn_CalendarTbl
/************************************************
Function: fn_CalendarTbl
Purpose: Creates an on the fly table for date 
  searches and joins 
Returns: A table of one year of dates
Example Call: select * from fn_CalendarTbl(2003)
  select * from fn_CalendarTbl(2004)  
Notes:  Accounts for leap years  
Authors: Anatoly Lubarsky / Thomas Darmstandler
--------        ---------      ----     ----------
History: 4/8/2004  Created
************************************************/

(@yr int)
returns @t2 table (dt datetime)

as
begin
declare @startDt datetime,
 @dt datetime,
 @leapDt datetime,
 @retVal int,
 @enterDt varchar(20)

declare @t table (dt smalldatetime)

set @startDt = convert(smalldatetime, 
               convert(varchar(4), @yr) + '0101')

insert into @t
select dt
from
 (select
 @startDt +
 n3.num * 100 +
 n2.num * 10 +
 n1.num as dt
 from
  (select 0 as num union all
  select 1 union all
  select 2 union all
  select 3 union all
  select 4 union all
  select 5 union all
  select 6 union all
  select 7 union all
  select 8 union all
  select 9
  ) n1,
   (select 0 as num union all
   select 1 union all
   select 2 union all
   select 3 union all
   select 4 union all
   select 5 union all
   select 6 union all
   select 7 union all
   select 8 union all
   select 9) n2,
    (select 0 as num union all
    select 1 union all
    select 2 union all
    select 3) n3) gencalendar
order by 1

---------------
-- accounts for leap years
set @enterDt ='2/28/' + cast(@yr as varchar(4))
set @leapDt = (select dateadd(d,1,@enterDt))
set @retVal = (select datepart(d,@leapDt))

if @retVal = 29
  begin
 insert into @t2
 select top 366 * 
 from @t
  end 
else
  begin

 insert into @t2
 select top 365 * 
 from @t
  end 

return
end 


My response: It works fine, but the perfomance is not good because:


1) performing the whole insert twice
2) extra variables for example: extra table variable, 
extra datetime variable. In udf you have to be aware of it.
3) datetime conversion formats won't work under certain locals 


So I enhanced my version #1 a bit. So you get version #2:


/*
########################################################
#
#   F_TBL_CALENDAR
#
#   version #2
#
#   use: select * from dbo.F_TBL_CALENDAR(2003)
#
########################################################
*/

alter function F_TBL_CALENDAR
(@p_year smallint)
returns @tbl table (cal_date smalldatetime)
as
begin
 
   -- 1 --
   -- check for leap year   
   declare @p_leap_date smalldatetime
   declare @p_check_day int
 
   set @p_leap_date = cast(@p_year as varchar(4)) + '0228'
   set @p_leap_date = (select dateadd(d, 1, @p_leap_date))
   set @p_check_day = (select datepart(d, @p_leap_date))

   -- 2 --
   -- create start date
   declare @p_start_date smalldatetime
   set @p_start_date = convert(smalldatetime, 
                       convert(varchar(4), @p_year)
                       + '0101')
   -- 3 --
   -- generate dates
   insert into @tbl
   select top 365 cal_date
   from
   (
      select
         @p_start_date +
         n3.num * 100 +
         n2.num * 10 +
         n1.num as cal_date
      from
      (
         select 0 as num union all
         select 1 union all
         select 2 union all
         select 3 union all
         select 4 union all
         select 5 union all
         select 6 union all
         select 7 union all
         select 8 union all
         select 9
      ) n1,
      (
         select 0 as num union all
         select 1 union all
         select 2 union all
         select 3 union all
         select 4 union all
         select 5 union all
         select 6 union all
         select 7 union all
         select 8 union all
         select 9
      ) n2,
      (
         select 0 as num union all
         select 1 union all
         select 2 union all
         select 3
      ) n3
   ) gencalendar
   order by 1   
   
   -- 4 --
   -- if it was a leap year
   -- add 31/12
   if (@p_check_day = 29)  
      insert into @tbl values(convert(smalldatetime, 
                                      convert(varchar(4), 
                                      @p_year)
                                      + '1231'))   
   return
end

:)

Related Posts:

Saturday, April 10, 2004 4:14 AM

Comments

# re: Generate Calendar Dates.Final Version
/*
########################################################
#
# F_TBL_CALENDAR
#
# version #2
#
# use: select * from dbo.udf_year_dates_table(2004)
#
########################################################
*/

create function udf_year_dates_table (@year int)
returns @year_dates table(date1 datetime) as
begin
declare @cnt int, @start_date datetime
select @start_date = convert(datetime,'01/01/'+convert(varchar(4),@year),120)
select @cnt = datediff(dd,@start_date ,dateadd(yy,1,@start_date))
insert into @year_dates
select
@start_date +
n3.num * 100 +
n2.num * 10 +
n1.num as date1
from
(
select 0 as num union all
select 1 union all
select 2 union all
select 3 union all
select 4 union all
select 5 union all
select 6 union all
select 7 union all
select 8 union all
select 9
) n1,
(
select 0 as num union all
select 1 union all
select 2 union all
select 3 union all
select 4 union all
select 5 union all
select 6 union all
select 7 union all
select 8 union all
select 9
) n2,
(
select 0 as num union all
select 1 union all
select 2 union all
select 3
) n3
where
n3.num * 100 +
n2.num * 10 +
n1.num < @cnt
order by
@start_date +
n3.num * 100 +
n2.num * 10 +
n1.num

RETURN
end

4/12/2004 5:23 PM by Enigma

# re: Generate Calendar Dates. Version #2.
This function returns the dates for only the particular year you ask for ... no need to do a top 365 or top 366

4/12/2004 5:24 PM by Enigma

# re: Generate Calendar Dates. Version #2.
You don't need to do a top 365/366 using my function in version #2 too.

4/12/2004 11:25 PM by Anatoly Lubarsky

# re: Generate Calendar Dates. Version #2.
thanks for that:
select @cnt = datediff(dd,@start_date ,dateadd(yy,1,@start_date))

4/12/2004 11:35 PM by Anatoly Lubarsky

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