我如何在不包含的情况下将变量传递给另一个PHP文件？

Question

how would I pass the variable $u to another php file without using include or anything similar? Am I right in thinking that the u is fetched ($_GET) from the url?

if(isset($_GET["u"])){
   $u = preg_replace('#[^a-z0-9]#i', '', $_GET['u']);
}

Answer 1

Yes, you right. For your purposes you need to use $_SESSION and session mechanism or cookies.

Answer 2

You could pass it via GET , POST or sessions .

GET:

file1.php

<?php
$u = 'something';
echo '<a href="http://website.com/file2.php?u=' . $u . '">Click me!</a>';
?>

file2.php

<?php
if(isset($_GET['u'])) {
    echo $_GET['u'];
}
?>

Sessions:

file1.php

<?php
session_start();
if(!isset($_SESSION['u'])) {
    $_SESSION['u'] = 'something';
}
?>

file2.php

<?php
session_start();
if(isset($_SESSION['u'])) {
    echo $_SESSION['u'];
}
?>

Answer 3

Yes, the 'u' is fetched from the url. eg http://domain.tld/index.php?u=bob will produce 'bob' as 'u'. Add u=whatever to the url and the 'u' will be on that page or if you don't want to do that you will need to store it in a session and/or cookie.