Generating random number error [closed] - php

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I am testing a php function i found on the php documentation based on this comment:
http://www.php.net/manual/en/function.rand.php#108861
<?php
function RandNumber($e) {
for ($i = 0; $i < $e; $i++) {
$rand = $rand . rand(0, 9);
}
return $rand;
}
echo RandNumber(4);
// Outputs a 6 digit random number
?>
I get the error Notice:
Undefined variable: rand in /var/www/eod.php on line 7
This is line 7:
$rand = $rand . rand(0, 9);
Why is causing this error since the function works as expected?

Answer 1

Initialize your $rand variable to remove this warning :
function RandNumber($e){
$rand="";
for($i=0;$i<$e;$i++){
$rand = $rand.rand(0, 9);
}
return $rand;
}
Note that you may simplify your code :
function RandNumber($e){
$rand="";
for(;$e-->0;){ // no need for an additional variable
$rand .= rand(0, 9); // addition and assignement with one operator
}
return $rand;
}

Answer 2

To output a random 6 digit number (If you need larger numbers, greater than mt_getrandmax(), this solution would fail), simply use
mt_rand(100000, 999999);
To include numbers with leading zeros (000123), you can use a combination of my_rand(0, 999999) and str_pad($number, 6, "0", STR_PAD_LEFT);
No need for fancy looping.

Answer 3

This is not an error.. it's a notice message.. If you don't want to appear that notice you should define it before doing the loop:
$rand = '';
for($i=0;$i<$e;$i++){
$rand = $rand . rand(0, 9);
}