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Writing Functions in T-SQL

iolike
iolike asked
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Last Modified: 2012-05-06
I am having problems specifying the date argument in this question. Any idea

Write a function named twoDigitYearPart that accepts a datetime as an argument and returns the last two digits of the year as a char(2) type.
 

CREATE FUNCTION dbo.twoDigitYearPart (
 
@date date )
 
Returns Char (2)
 
as
 
BEGIN
 
RETURN (SELECT CASE WHEN @date THEN @date = @date char(2) END)
 
end

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AneeshDatabase Consultant
CERTIFIED EXPERT
Top Expert 2009

Commented:
RETURN (SELECT YEAR(@Date) )

Author

Commented:
when I  execute the following,  I don't get 08. Any suggestions

SELECT dbo.twoDigitYearPart (2008)
The YEAR function returns an integer.

You can specify a format for the date using the CONVERT function.  (101 is MM/DD/YYYY)

Try this...
CREATE FUNCTION dbo.twoDigitYearPart(@date datetime) 
 
RETURNS char(2) 
 
AS 
 
BEGIN
 
	DECLARE @output char(2) 
	SET @output = RIGHT(CONVERT(varchar, @date, 101), 2) 
	RETURN @output 
 
END 
 
GO 

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Author

Commented:
Althoug I execute SELECT dbo.twoDigitYearPart5 (2008), I get 05 as result.  It returns 05 regardless what year I execute
Database Consultant
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Top Expert 2009
Commented:
This one is on us!
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I wouldn't have expected the SQL Server to accept an integer.  Odd...

The SQL Server seems to be returning the date representing the number of days since 1/1/1900 (for 2008 that's 6/2/1905, so the function returns '05').  You should pass a datetime value instead.  (i.e. SELECT dbo.twoDigitYearPart('1/1/2008') )

Here, you can see the behavior of the SQL Server for yourself:
PRINT CONVERT(datetime, 0) 
PRINT CONVERT(datetime, 2008) 
PRINT dbo.twoDigitYearPart(GETDATE()) 

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