Can Sql Server generate a unique 13 digit number - sql

Question

I need to generate a unique 13 digit number.
Can Sql Server generate this number for me somehow if I create a table with the 13 digit number as a primary key?
Update
I want the number to look like a random number, so not an autoincrementing number.
It has to be 13 digits, and it shouldn't be auto-incrementing, and it should be unique. The number shouldn't have many zero's in it but it can contain numbers from 0-9.
This number should look like a credit card number, so no trailing zeros.

Answer 1

My suggestion would be to have an identity column on the table that auto-increments. Then, define your value based on this. A simple way would be:
create table t (
tId int identity(1, 1) not null,
. . .
myId cast(rand(tId)*10000000000000 as varchar(13))
)
This shows it as a computed column. Of course, you can assign the value when each row is created. This is not guaranteed to produce different results, but it is highly, highly unlikely that you would see a collision.
The following is an alternative is also not guaranteed, but might work:
create table t (
tId varchar(13) default cast(cast(rand(checksum(getdate())*10000000000000 as bigint) as varchar(13)
. . .
)
EDIT:
The chance of a collision is a bit higher than I expected -- my intuition on 13-digit hash codes is, I guess, not what it should be.
In any case, there are two sources of collisions. The first is the random number generator producing the same value. To handle that, just make the assumption that the random number generator in conjunction with checksum() really is random. So, the question is: What is the chance of two random numbers less than 10,000,000,000,000 being the same value? I'll let interested parties search the web for a formula to calculate this.
If you generate 1,000 numbers. Then the probability is basically 0% that any two would be the same. That is, you are safe for the first 1000 numbers, if you assume they are distinct. Here is a summary:
1,000 0.0000%
10,000 0.0005%
100,000 0.0500%
1,000,000 4.8771%
10,000,000 99.3262%
Up to a few hundred thousand values, you are probably pretty safe. When you get into the millions -- even the low millions -- the chance of collision increases substantially.
At some point, if you want lots and lots of unique values, you are going to have to create a table that contains the unique values and a process for choosing a value not in the table.
As John Barça points out: Do not use this method for cat photos on Facebook.

Answer 2

Just create a Guid (select newid() ) & parse it... remove the { and the '-' & do a length of 13 select on it.

Answer 3

I use a similar method to generate random/unique table names in a reporting system. It might do the trick for you. Just adjust the multiplier to impact the final integer length.
SELECT CONVERT(BIGINT,RAND()*10000000000000)
As a table...
CREATE TABLE #test (testID INT,
UniqueID AS CONVERT(BIGINT,RAND()*10000000000000))
INSERT INTO #test (testID)
SELECT 1
SELECT * FROM #TEST
DROP TABLE #test
Just insert a value into testID (can be 1 every time) and a new UniqueID will generate. You should have a primary key on any production table though.
NB: While the chances of a duplicate ever happening are very small, it could still happen.

Answer 4

SELECT CEILING(RAND()*9999999999999)

Answer 5

something like this might work:
use [chamomile];
go
if object_id(N'[utility].[table_01]', N'U') is not null
drop table [utility].[table_01];
go
if object_id(N'[utility].[generate_random_sequence]', N'FN') is not null
drop function [utility].[generate_random_sequence];
go
/*
select [utility].[generate_random_sequence] (rand());
*/
create function [utility].[generate_random_sequence] (
#random [float])
returns [bigint]
as
begin
declare #return [bigint] = ceiling(#random * 9999999999999);
while #return > 9999999999999
or #return < 1000000000000
set #return = ceiling(#random * 9999999999999);
return #return;
end;
go
if object_id(N'[utility].[table_01]', N'U') is not null
drop table [utility].[table_01];
go
create table [utility].[table_01] (
[my_id] as [utility].[generate_random_sequence] (rand())
, [flower] [sysname]
);
go
insert into [utility].[table_01]
([flower])
values (N'rose');
select *
from [utility].[table_01];