Call variable outside while loop

Question

I have a simple code here:

$rand     = mt_rand(100000, 999999);
$sid      = array(
    ":sponsorID" => $rand
);
$accounts = $db->select("accounts", "sponsorID = :sponsorID", $sid);
while (count($accounts) > 0) {

    $new = mt_rand(100000, 999999);
}
echo $new;

What it basically does is to check if the generated number from $rand is already in the database and will set a $new variable. My problem is when I echo out $new, I always get an undefined index of that variable. How can I call that variable outside the while loop?

What is the sense of this? Why would you overwrite that variable $new in all iterations? I assume you want to add a break or continue statement to that loop... Also: this will be an endless loop once there is an entry in the database, since you do not modify $accounts inside the loop... — arkascha
– arkascha, Commented Aug 2, 2015 at 6:46
@arkascha thank you for pointing that out.. I forgot that. Should I use continue because I want it to go back to the condition when it is already present in the database — wobsoriano
– wobsoriano, Commented Aug 2, 2015 at 6:49
I could imagine your want to use a for or a foreach loop instead of that while... — arkascha
– arkascha, Commented Aug 2, 2015 at 6:57

Mureinik · Accepted Answer · 2015-08-02 06:43:45Z

3

You are only defining $new inside the loop, so it's undefined outside it. Just declare it before the loop and you should be fine:

$new = NULL;
while (count($accounts) > 0) {
    $new = mt_rand(100000, 999999);
}
echo $new;

answered Aug 2, 2015 at 6:43

Mureinik

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1 Comment

wobsoriano Over a year ago

I always get null with this

arkascha · Accepted Answer · 2015-08-02 07:36:26Z

1

There are a few issues with your code. This is a modified version that outputs what you are looking for:

<?php 

$sid = array(
  ":sponsorID" => $rand
);
$accounts = $db->select("accounts", "sponsorID = :sponsorID", $sid);

$new = null; 
do { 
  $new = mt_rand(100000, 999999); 
} while (in_array($new, array_column($accounts, 'sponsorId'))); 

echo $new;

Note: this requires at least php 5.5 because of the array_column() call. But that makes the code simple and easy to understand.

edited Aug 2, 2015 at 7:36

answered Aug 2, 2015 at 7:35

arkascha

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