asked on

PHP Programming Problem

I have a randomly assigned variable named $rand. I also have a set variable named $db. If $rand eqauls $db, I would like the script below to regenerate $rand up to seven times before displaying the "Stopped trying error." Why won't this code work? Attached is the code:

$rand = rand(100000, 999999);
$db = 5555555;
if ($rand != $db) { 
	echo $rand;
} else {
	for ($counter = 1; $counter < 7; $counter++) {
	$rand = rand(100000, 999999);
	if ($rand != $db) {
		echo $rand;
		} else { 
		echo "Random number could not be generated. Still trying... ";
		if ($counter == 7) {
		echo "Stopped trying."; }
	}
}

Open in new window

ASKER CERTIFIED SOLUTION

glcummins

membership

This solution is only available to members.

To access this solution, you must be a member of Experts Exchange.

Start Free Trial

EMB01

ASKER

Thanks, didn't see that - one problem remains though - when I take control of both variables to test the for function, it seems it doesn't echo the "Stopped trying" text. Why is this?

<?php
$rand = 555555;
$db = 5555555;
if ($rand != $db)
{ 
	echo $rand;
}
else
{
	for ($counter = 1; $counter < 7; $counter++)
	{
		$rand = 555555;
		if ($rand != $db)
		{
			echo $rand;
		}
		else
		{ 
			echo "Random number could not be generated. Still trying...\n ";
			if ($counter == 7)
			{
				echo "Stopped trying.\n";
			}
		}
	}
}
?>

Open in new window

glcummins

$counter will never equal 7 based on your for() statement. The loop only executes for $counter = 1 through $counter = 6. Less than seven will never equal seven.

EMB01

ASKER

Well, I thought that, too. Then I tried 6 and it didn't work either.

glcummins

$rand and $db also have to be equal. That block will never execute if they are unequal.

SOLUTION

glcummins

membership

This solution is only available to members.

To access this solution, you must be a member of Experts Exchange.

Start Free Trial

SOLUTION