从URL获取变量

i am new to php, but im trying. i need you guys help. i have the following url in the browser address bar www.dome.com\mypage.php?stu=12234342 i am trying to pass the url from the main page to the select case page call select.php if i should echo the url i get www.dome.com\select.php. so i have decided to echo $_SERVER['HTTP_REFERER'] instead, this gives me the correct url. how can i echo the variable from www.dome.com\mypage.php?stu=12234342 (12234342) unto select.php. select.php contains code that needs the $var stu=12234342 in order to display the correct message.

$request_url=$_SERVER['HTTP_REFERER'] ; // takes the url from the browers echo $request_url;

$cOption = $_GET['id'];

switch($cOption) {
    case 1:
    echo ' some text';
        break;
    case 2:
    echo ' this page.php';
        break;

    case 3:
    echo 'got it'; 
        break;
    default:
        echo 'Whoops, didn\'t understand that option: <i>'.$cOption.'</i>';
}

?>

try

echo $_GET['stu'];

on select.php

That's why you need to call the select.php file like this: www.dome.com/select.php?stu=12234342

and then you can add:

echo $_GET['stu'];

By the way, you need to research about XSS, because that's a huge vulnerability.

You may use parse_url() and parse_string() to grab the variable from a url:

<?php 

//assuming www.dome.com/mypage.php?stu=12234342;
  $url=$_SERVER['HTTP_REFERER'];

//parse the url to get the query_string-part
  $parsed_url=parse_url($url);

//create variables from the query_string
  parse_str($parsed_url['query'], $unsafe_vars);

//use the variables
  echo $unsafe_vars['stu'];//outputs 12234342
?>

But note: you can't rely on the availability of HTTP_REFERER.