Prairie Coder

Here's a function that will return the military julian equivalent of any date passed to it. The default is to return the current day's.

 

set ANSI_NULLS ON
set QUOTED_IDENTIFIER ON
go

-- =============================================
-- Description: Function to return Military Julian in the format of YYDDD
-- RETURNS:  varchar data type
-- INPUTS:  date or default
-- Example Using Today's Date: select dbo.fnGetMilitaryJulian(default)
-- Example Date: select dbo.fnGetMilitaryJulian('01/08/2008')
-- Example Leap Year: select dbo.fnGetMilitaryJulian('12/31/2012')
-- =============================================
ALTER FUNCTION [dbo].[fnGetMilitaryJulian]
(
 -- Add the parameters for the function here
 @theDate datetime = null
)
RETURNS varchar(5)
AS
BEGIN
 -- Declare the ordinal part of the variable
 DECLARE @miljul varchar(3)

--treat an empty string or spaces as a null
if rtrim(ltrim(@theDate)) = ''
 set @theDate = NULL

DECLARE @d varchar(2);
DECLARE @m varchar(2);
DECLARE @y int;

DECLARE @firstofyear datetime;

DECLARE @yearpart varchar(2);
DECLARE @tempjul varchar(5);

set @m = 01;
set @d = 01;

-- set year part on the date passed in
if @theDate is not null
  BEGIN
   set @y = year(@theDate);
 set @yearpart = RIGHT(convert(varchar(4),year(@theDate)),2);
  END
else
  BEGIN
 -- set year part for today
 set @y = year(getdate());
 set @yearpart = RIGHT(convert(varchar(4),year(getdate())),2);
  END
 -- finally determine the first day of the year requested
 select @firstofyear = convert(datetime,@m +'-'+ @d +'-'+ convert(varchar,@y))

 -- get the ordinal day from the constructed date
 if @theDate is not null
   BEGIN
     SELECT @miljul = convert(varchar(3),datediff(day,@firstofyear, @theDate) + 1)
   END
 else
   BEGIN
  -- set miljul for today
  SELECT @miljul = convert(varchar(3),datediff(day,@firstofyear, getdate()) + 1)
   END

 -- insure the ordinal day is 3 digits long
 set @miljul = right(replicate('0',3) + @miljul,3)

RETURN @yearpart + @miljul

END