GET URL parameter in PHP

Question

$_GET is not a function or language construct—it's just a variable (an array). Try:

<?php
echo $_GET['link'];

In particular, it's a superglobal: a built-in variable that's populated by PHP and is available in all scopes (you can use it from inside a function without the global keyword).

Since the variable might not exist, you could (and should) ensure your code does not trigger notices with:

<?php
if (isset($_GET['link'])) {
    echo $_GET['link'];
} else {
    // Fallback behaviour goes here
}

Alternatively, if you want to skip manual index checks and maybe add further validations you can use the filter extension:

<?php
echo filter_input(INPUT_GET, 'link', FILTER_SANITIZE_URL);

Last but not least, you can use the null coalescing operator (available since PHP/7.0) to handle missing parameters:

echo $_GET['link'] ?? 'Fallback value';

Please post your code,

<?php
    echo $_GET['link'];
?>

or

<?php
    echo $_REQUEST['link'];
?>

do work...

Use this:

$parameter = $_SERVER['QUERY_STRING'];
echo $parameter;

Or just use:

$parameter = $_GET['link'];
echo $parameter ;

To make sure you're always on the safe side, without getting all kinds of unwanted code insertion use FILTERS:

echo filter_input(INPUT_GET,"link",FILTER_SANITIZE_STRING);

More reading on php.net function filter_input, or check out the description of the different filters

The accepted answer is good. But if you have a scenario like this:

http://www.mydomain.me/index.php?state=California.php#Berkeley

You can treat the named anchor as a query string like this:

http://www.mydomain.me/index.php?state=California.php&city=Berkeley

Then, access it like this:

$Url = $_GET['state']."#".$_GET['city'];

$Query_String  = explode("&", explode("?", $_SERVER['REQUEST_URI'])[1] );
var_dump($Query_String)

Array ( [ 0] => link=www.google.com )