How to generate random integer not present in SQL database? - php

Question

Currently working on a login script that would allow for multiple users with the same username to exist. The current plan is to generate a random "secret" user id that will take the place of the actual username. So how would I go about generating a random integer and checking to see if has been added?
This is my attempt at the code; however it does not seem to work as it does not seem to do anything.
$looptime = 100;
while ($looptime > 0) {
$userid = rand(0, 999999999);
$SQL = "SELECT * FROM Accounts WHERE username = '" . $userid . "'";
$result_id = #mysql_query($SQL) or die("DATABASE ERROR!");
$total = mysql_num_rows($result_id);
if (!$total) {
$looping = 0;
//CREATE THE ACCOUNT
mysql_query("INSERT INTO Accounts (id,username,password, email, money, activated) VALUES ('', '$username','$password', '$email', '0', '1')") or die("REGISTER ERROR!"); //1=account activated
//CONNECTED
echo "IMCONNECTED";
}
$looptime--;
}
EDIT: The code/number should be fairly easy to remember/jot down somewhere as the user will be able to view it and/or jot it down for account recovery purposes.

Answer 1

I would suggest either using a UUID/GUID (not an integer) to minimize the possibility of clashes, or using an identity column (not random) to guarantee uniqueness. Does it have to be an integer and does it have to be random?

Answer 2

Are you using an integer for the ID in the table? You could append this ID to the username. For example: MyUsername1234, MyUsername1245.

Answer 3

Here is a way you could do it. Create a scalar-variable function in your database (similar to below):
CREATE FUNCTION [dbo].[fn_randomNumber] (#guid nvarchar(128))
RETURNS int AS
BEGIN
SET #guid = REPLACE(#guid, '-', '');
DECLARE #idInt varchar(Max) = '';
DECLARE #i INT = 0;
DECLARE #char VARCHAR;
WHILE(#i < LEN(#guid))
BEGIN
SET #char = (SUBSTRING(#guid, #i, 1));
IF(ISNUMERIC(#char) = 1)
BEGIN
SET #idInt = #idInt + #char;
IF(LEN(#idInt) = 9)
BEGIN
BREAK;
END
END
SET #i = #i + 1;
END
RETURN CAST(#idInt as INT);
END
GO
Then execute this script:
SELECT [dbo].[fn_randomNumber] (
newid())
Of course you will want to evaluate the result to make sure it doesn't already exist.