How to make UDF deterministic

I have a user defined function in SQL Server 2000 that splits a Name field into First, Last, Middle names. Originally I utilized the CHARINDEX function to find the spaces in the string. Since I then found out that function is by definition Non-deterministic that didn't work for me, so I re-wrote the function a bit differently.

I am using the function as a field's formula in my table structure. I need the ability to create an index on that field so need the function to be deterministic. Apparently there is still something about how I designed the function that causes it to be non-deterministic. The function code is below; if anyone can help point me in the right direction it'd be greatly appreciated.

ALTER function fSplitName_test
(
@.name varchar(100),
@.fname int)
returns varchar(55)
as

begin
declare @.Fname_position int, @.Lname_position int, @.Mname_position int, @.SplitName varchar(55), @.iint int, @.iLen int,
@.sChar char(1), @.sName varchar(55)

set @.name = ltrim(replace(@.name, ' JR', ''))
set @.sName = @.name
set @.ilen = len(rtrim(@.name))
set @.Fname_position = 0
set @.Lname_position = 0
set @.iint = 0

while @.iint <> @.ilen
begin
set @.schar = left(@.sName, 1)
begin
if @.schar = ' '
begin
if @.Fname_position = 0
set @.Fname_position = @.iint
else
set @.Lname_position = @.iint
end
set @.iint = @.iint + 1
set @.sName = right(@.sName, @.ilen - @.iint)
end
end
if @.fname = 1
begin
if @.fname_position <> 0
set @.SplitName = left(@.name, @.Fname_position)
else
set @.splitname = 'unkn'
end
else
begin
if @.fname = 2
begin
if @.lname_position <> 0
set @.SplitName = right(@.name, @.ilen - @.Lname_position-1)
else
set @.SplitName = right(@.name, @.ilen - @.fname_position)
end
else
begin
if @.fname_position + @.lname_position <> @.ilen and @.lname_position <> 0
begin
set @.SplitName = left(@.name, @.lname_position)
set @.splitname = right(@.splitname,len(@.splitname) - @.fname_position - 1)
end
else
set @.splitname = 'unkn'
end
end
return @.SplitName
endYou might get more response if you post this under the "Microsoft SQL Server" forum. This one's for generic SQL questions - I for one know nothing about SQL Server.|||what if a person has a lastname like this "DELA CRUZ"
can you post an example of the name parameter ex: "Michael Angelo Rodriguez" or "Michael Angelo Dela Cruz" coz as you can see on the 2 ex. it is hard to detemine the first name on the first example and last name on the latter the name should have a middle initial or comma perhaps.|||This is probably completely wrong, but I wouldn't put it past SQL Server to think that you were trying to modify the parameters. That would, of course, make the function non-deterministic. Try copying them to local variables.

Otherwise, the way to debug a problem like this is to break your function into small functions and verify that the small functions are all deterministic.sql

How to locate last occurrence of character in string

charindex() function lets you find first occurrence of a particular
character in a string.
Is there similar function to find last occurrence of character in
string.
I could use reverse() or some script to find that, but it will be
inefficient.
Could anybody suggest me efficient solution for this.
Thanks and Regards,
AvinAvi, please do not multipost as I have already answered in.programming
group.
"Avi" <avimandale@.gmail.com> wrote in message
news:271710bf-a8dd-4a82-8bc2-076b920d6919@.k39g2000hsf.googlegroups.com...
> charindex() function lets you find first occurrence of a particular
> character in a string.
> Is there similar function to find last occurrence of character in
> string.
> I could use reverse() or some script to find that, but it will be
> inefficient.
> Could anybody suggest me efficient solution for this.
> Thanks and Regards,
> Avin|||On Jan 28, 5:25=A0pm, Avi <avimand...@.gmail.com> wrote:
> charindex() function lets you find first occurrence of a particular
> character in a string.
> Is there similar function to find last occurrence of character in
> string.
> I could use reverse() or some script to find that, but it will be
> inefficient.
> Could anybody suggest me efficient solution for this.
> Thanks and Regards,
> Avin
Hi Avin,
declare @.string1 varchar(50), @.string2 varchar(50)
select @.string1 =3D 'abc', @.string2=3D 'oooabcooabco'
select (datalength(@.string2)-1)-
charindex(reverse(@.string1),reverse(@.string2))
HTH