I was looking through some of the active threads on SQLServerCentral.com tonight, and I ran across one with some code that looked familiar. While reading this thread, I noticed that the function returned a fixed length random character string. This told me that if you needed a random character string of a different length, you'd need another function.
How can you get around this? You can't use the function NEWID() in a function, SQL Server 2005 doesn't like it. There is a way to do this. Create a view that returns the result of the function NEWID(), and use that in your function to return a random string of different lengths.
Here is the code I put together:
create view dbo.MyNewID asselect newid() as NewIDValue;
go
create function dbo.ufn_RandomString(
@pStringLength int = 20
) returns varchar(max)
as begin
declare @RandomString varchar(max);
with
a1 as (select 1 as N union all
select 1 union all
select 1 union all
select 1 union all
select 1 union all
select 1 union all
select 1 union all
select 1 union all
select 1 union all
select 1),
a2 as (select
1 as N
from
a1 as a
cross join a1 as b),
a3 as (select
1 as N
from
a2 as a
cross join a2 as b),
a4 as (select
1 as N
from
a3 as a
cross join a2 as b),
Tally as (select
row_number() over (order by N) as N
from
a4)
, cteRandomString (
RandomString
) as (
select top (@pStringLength)
substring(x,(abs(checksum((select NewIDValue from MyNewID)))%36)+1,1)
from
Tally cross join (select x='0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ') a
)
select @RandomString =
replace((select
',' + RandomString
from
cteRandomString
for xml path ('')),',','');
return (@RandomString);
end
go