PHP 4.0.6 Question

[Anon]

Hello

im using PHP 4.0.6 And if(isset($_GET['fee']))
dosnt seem to work. does anyone know how i can do this?

this is what i got
http://url.com/index.php?order=true&18=gum&19=coke

so i want to print the value of 18 & 19
gum & coke/ 🙂

here is my code..

<?php
if ($order){
$count = 0;
$data = file($dbFile);
$total = count($data);
for ($i = 0; $i <= $total; $i++)
{
$wantedLine = $i;
$lineData = $data[$wantedLine];
$lineData = trim($lineData);
$lineData = explode("|", $lineData);
$count++;
echo "gum and coke here?";
}
}
?>

[Anon]

For that version of php
if(isset($fee))

A valid variable name starts with a letter or underscore there for 18 and 19 will not work.

[Anon]

Either use $HTTP_GET_VARS['fee'] instead of $_GET['fee'] or use PHP version >= 4.1 instead of PHP 4.0.6 🙂