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vovka

Valued Member

Points: 63
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June 3, 2003 at 7:48 pm

#101506

Hi-, I have really weird problem.
My biz logic request to generate unique id, so the sp gets parametr "type" varchar(3), next this id have to be started with 'A' pls "type" pls random number to fill up the rest in total 9 char.
Here's the sample of rand() function I'm using to get 5 char number...
declare @rnd int,
@Upper int,
@Lower int
set @Lower = (DATEPART(ss, GETDATE()) * 1 )
set @upper = (DATEPART(ms, GETDATE()) * 100 )
SET @rnd = CAST(((@Upper - @Lower + 1) * RAND() + @Lower) AS Int)..
after that there is a cast to varchar and len check and concatination all 3 in one that goes to insert. All works fine untill I do isert which works fine when I'm in debugging stand alone mode.
When I run this sp with real data my id comes out as 'A*'.? Appreciate your ideas!!
Thnx.
PS. All happens in .NET SQL2000/SP3

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Antares686 SSC Guru Points: 125444 More actions · Answer 1

I have no issues myself. Can you provide more code to the point of where the A number is finally generated?

vovka Valued Member Points: 63 More actions · Answer 2

quote:
I have no issues myself. Can you provide more code to the point of where the A number is finally generated?
OK. Here is the rest of it. Hope it helps..
SET @rnd2 = CAST(@rnd as varchar(5))
if @type is null
Begin
Set @aimid = 'A099' + @rnd2
Select @len = len(@aimid)
if @len < 9
Begin
if @len = 5
Set @rnd = @rnd * 10000
else
if @len = 6
Set @rnd = @rnd * 1000
else
if @len = 7
Set @rnd = @rnd * 100
else
if @len = 8
Set @rnd = @rnd * 10
SET @rnd2 = CAST(@rnd as varchar(5))
Set @aimid = ''
Set @aimid = 'A099' + @rnd2

vovka Valued Member Points: 63 More actions · Answer 3

OK. I thinh I know what's wrong. The middle string get lost if it hardcoded in concat.

I passed it as parm and it worked OK.

Thnx.