User defined function exercises

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I need to create a user defined function to find simple Interest, with Principal amount, no. of years and rate as parameters.
The next one is to create a user defined function to find the factorial of a number.
Help me out...

Posted 8-Oct-12 2:42am

bolshie6

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Comments

Tim Corey 8-Oct-12 8:47am

"Help me out" isn't exactly a question. We aren't here to do your work. What have you tried? Do you know what a UDF is? Have you figured out how to do a query for this? Sorry, right now this looks to be a "do my work for me" post and I don't respond well to that.

Herman<T>.Instance 8-Oct-12 8:58am

We can give him a google search key like this one.

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Santhosh Kumar Jayaraman · Accepted Answer · 2012-10-08T03:03:00

Solution 1

SQL

create FUNCTION dbo.Factorial ( @iNumber bigint )
RETURNS INT
AS
BEGIN
DECLARE @i  bigint

    IF @iNumber <= 1
        SET @i = 1
    ELSE
        SET @i = @iNumber * dbo.Factorial( @iNumber - 1 )
RETURN (@i)
END


declare @int bigint
exec @int= dbo.Factorial 12
select @int

Posted 8-Oct-12 3:03am

Santhosh Kumar Jayaraman

Comments

bolshie6 9-Oct-12 1:10am

This is my UDF for finding Simple interest. Theres always an error near RETURN

CREATE FUNCTION ufn_SimpleInterest (@Principal NUMERIC(19,4), @TimePeriod NUMERIC(19,2), @Rate NUMERIC(19,9)
RETURN NUMERIC(19,4)
AS
BEGIN
DECLARE @SimpleInterest NUMERIC(19,4)
SELECT @SimpleInterest = ( @Principal * @TimePeriod * @Rate) / 100
RETURN @SimpleInterest
END

Santhosh Kumar Jayaraman 9-Oct-12 1:11am

try this
CREATE FUNCTION ufn_SimpleInterest (@Principal NUMERIC(19,4), @TimePeriod NUMERIC(19,2), @Rate NUMERIC(19,9))
RETURNs INT
AS
BEGIN
DECLARE @SimpleInterest NUMERIC(19,4)
SELECT @SimpleInterest = ( @Principal * @TimePeriod * @Rate) / 100
RETURN @SimpleInterest
END

bolshie6 9-Oct-12 1:18am

Thank you, It works. I've noted that its 'RETURNS'.

Santhosh Kumar Jayaraman 9-Oct-12 1:20am

also u missed closing bracket next to @Rate NUMERIC(19,9)