php - for loop inside a while loop, correct syntax?

Question

I have this loop

while (count($arr) < 7)
        {

            $string = 'xx';
            for ($i=0; strlen($string) < 4; $i++)
            {
                $string = $string.'1';
            }
            echo "<br>array length is less than 7, word $string is created.";
            $arr[] = $string;
        }

Every time I run this part of the code, my xampp local server would time out and gives the server not found error.

I have found that if I delete the inner for loop, it would run fine. Is there anything wrong with putting the for loop inside the while loop?

Also, my conditional statement strlen($string) < 4 in the for loop does not have any reference to the $i variable, but I don't see any thing not logical of having a conditional statement not related to the counter. Am I wrong, does it require some sort of comparison against the counter?

TIA

You never use $i, so you can change it to while( strlen($string) < 4) — nickb
– nickb, Commented Mar 13, 2012 at 0:28
Server not found error?? What's that got to do with php syntax? — Adam
– Adam, Commented Mar 13, 2012 at 0:28
I see absolutely nothing wrong with it.. I'll keep looking thought.. — DanRedux
– DanRedux, Commented Mar 13, 2012 at 0:28
at first glance it looks fine & nothing to do with nested loops. Personally I use NetBeans; you might choose Eclipse, Zend, etc. But, if you don't have a debugger to step through your code you are not going to get far. — Mawg
– Mawg, Commented Mar 13, 2012 at 0:29
Nick, she may want to use it later. Nothing wrong with it, unless you suspect it's causing the error somehow.. The only way it could be causing an error is if $string was assigned before to it by reference, which I doubt. $string=&$i would cause errors like this... — DanRedux
– DanRedux, Commented Mar 13, 2012 at 0:30

John3136 · Accepted Answer · 2012-03-13 00:30:53Z

1

Nothing wrong with having a for inside a while.

Your "for" would be better a bit clearer as

while(strlen($string) < 4) {
    $string = $string.'1';
}

answered Mar 13, 2012 at 0:30

John3136

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3 Comments

Jamex Over a year ago

Thanks John, the reason I have $i is because I am using $i as an array index, that part of the code was left out.

John3136 Over a year ago

Well, your PHP is valid and works for me - problem must be elsewhere (perhaps in some of the code you've omitted?)

Jamex Over a year ago

Thanks again, I am walking through my very short omitted code to see if there is any thing else. At least I know the error is not due to the loops.

iambriansreed · Accepted Answer · 2012-03-13 00:35:38Z

0

I don't see any issues whatsoever with your php.

I copied your code exactly and there was no infinite loop at all.

The result I got was as follows:

<br>array length is less than 7, word xx11 is created.
<br>array length is less than 7, word xx11 is created.
<br>array length is less than 7, word xx11 is created.
<br>array length is less than 7, word xx11 is created.
<br>array length is less than 7, word xx11 is created.
<br>array length is less than 7, word xx11 is created.
<br>array length is less than 7, word xx11 is created.

My only suggestion would be to change:

$string = $string.'1';

to

$string .= '1';

answered Mar 13, 2012 at 0:35

iambriansreed

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1 Comment

Jamex Over a year ago

Thanks Brian for testing the code, I edited the code for use as an example, not a real code.

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