stumped........

[31337_one]

I have apache 1.3xx with php 4.2.1 running windows 2000 Adv. Servr hosting my own domain.

my question is.. looking at the below script..

<?
$a=0;

print $a;

?>

this produces a very simple zero on the screen, no problem.. but when I try to load a variable in the URL exp.

www.mydomain.com/test.php?a=4

it does not produce a 4, but instead still will only show the 0.

is my code wrong? in either the script or URL? I do have my domain being forwarded from domainwizardry.com to my IP address..

this is a simple problem that will help me fix a larger more complex issue if I can get this figured out.

A friend of mine had told me that the variables "a=4" without quotes are not being "parsed"... I don't know what that means in regards to php scripting. He also said that my domainwizardry.com could be scrubbing the variables when forwared. But I don't think that true because i get the same results when I use localhost and the php file...

Any help in any direction would be greatly appreciated.

Thankx

😕

[edwardsbc]

By default, POST, GET, and COOKIE variables are not registered globally (as of PHP 4.x.x). You must retreive them from the super global arrays: $REQUEST (P,H,C), $POST, $GET, or $COOKIE.

$a=$_REQUEST['a'];
print $a;

[31337_one]

thanks...

I did try that but now I get

Notice: Undefined index: a in mydomain\test1.php on line 4

I am a mere novice with php so, am I not declaring something correctly now?

Thanks again.

[gutlass]

<?

if(!isset($a)) {
$a=0;
}

print $a;

?>

www.mydomain.com/test.php?a=4

will produce

"4"

[intenz]

The reason WHY it produced "0" instead of "4" is because you SET the $a variable as "0" after the URL tried to set $a as "4".

What you have to do is this...

if(!$a) $a = '0';
echo $a;
[code=php]
Good luck.