Request string without GET arguments [duplicate]

Question

This question already has answers here:

How to remove the querystring and get only the url? (16 answers)

Closed 5 months ago.

Is there a simple way to get the requested file or directory without the GET arguments? For example, if the URL is http://example.com/directory/file.php?paramater=value I would like to return just http://example.com/directory/file.php. I was surprised that there is not a simple index in $_SERVER[]. Did I miss one?

Answer 1

Edit: @T.Todua provided a newer answer to this question using parse_url.

(please upvote that answer so it can be more visible).

Edit2: Someone has been spamming and editing about extracting scheme, so I've added that at the bottom.

parse_url solution

The simplest solution would be:

echo parse_url($_SERVER["REQUEST_URI"], PHP_URL_PATH);

Parse_url is a built-in php function, who's sole purpose is to extract specific components from a url, including the PATH (everything before the first ?). As such, it is my new "best" solution to this problem.

strtok solution

Stackoverflow: How to remove the querystring and get only the url?

You can use strtok to get string before first occurence of ?

$url=strtok($_SERVER["REQUEST_URI"],'?');

Performance Note: This problem can also be solved using explode.

• Explode tends to perform better for cases splitting the sring only on a single delimiter.
• Strtok tends to perform better for cases utilizing multiple delimiters.

This application of strtok to return everything in a string before the first instance of a character will perform better than any other method in PHP, though WILL leave the querystring in memory.

An aside about Scheme (http/https) and $_SERVER vars

While OP did not ask about it, I suppose it is worth mentioning: parse_url should be used to extract any specific component from the url, please see the documentation for that function:

parse_url($actual_link, PHP_URL_SCHEME); 

Of note here, is that getting the full URL from a request is not a trivial task, and has many security implications. $_SERVER variables are your friend here, but they're a fickle friend, as apache/nginx configs, php environments, and even clients, can omit or alter these variables. All of this is well out of scope for this question, but it has been thoroughly discussed:

https://stackoverflow.com/a/6768831/1589379

It is important to note that these $_SERVER variables are populated at runtime, by whichever engine is doing the execution (/var/run/php/ or /etc/php/[version]/fpm/). These variables are passed from the OS, to the webserver (apache/nginx) to the php engine, and are modified and amended at each step. The only such variables that can be relied on are REQUEST_URI (because it's required by php), and those listed in RFC 3875 (see: PHP: $_SERVER ) because they are required of webservers.

please note: spaming links to your answers across other questions is not in good taste.

Answer 2

$uri_parts = explode('?', $_SERVER['REQUEST_URI'], 2);
$request_uri = $uri_parts[0];
echo $request_uri;

Answer 3

^{It's shocking how many of these upvoted/accepted answers are incomplete, so they don't answer the OP's question, after 7 years!}

• If you are on a page with URL like: http://example.com/directory/file.php?paramater=value

• ...and you would like to return just: http://example.com/directory/file.php

• then use:

  echo $_SERVER['REQUEST_SCHEME'].'://'.$_SERVER['SERVER_NAME'].$_SERVER['PHP_SELF'];

Answer 4

Here is a solution that takes into account different ports and https:

$pageURL = (@$_SERVER['HTTPS'] == 'on') ? 'https://' : 'http://';

if ($_SERVER['SERVER_PORT'] != '80')
  $pageURL .= $_SERVER['SERVER_NAME'].':'.$_SERVER['SERVER_PORT'].$_SERVER['PHP_SELF'];
else 
  $pageURL .= $_SERVER['SERVER_NAME'].$_SERVER['PHP_SELF'];  

Or a more basic solution that does not take other ports into account:

$pageURL = (@$_SERVER['HTTPS'] == 'on') ? 'https://' : 'http://';
$pageURL .= $_SERVER['SERVER_NAME'].$_SERVER['PHP_SELF']; 

Answer 5

You can use $_SERVER['REQUEST_URI'] to get requested path. Then, you'll need to remove the parameters...

$uri_parts = explode('?', $_SERVER['REQUEST_URI'], 2);

Then, add in the hostname and protocol.

echo 'http://' . $_SERVER['HTTP_HOST'] . $uri_parts[0];

You'll have to detect protocol as well, if you mix http: and https://. That I leave as an exercise for you. $_SERVER['REQUEST_SCHEME'] returns the protocol.

---

Putting it all together:

echo $_SERVER['REQUEST_SCHEME'] .'://'. $_SERVER['HTTP_HOST'] 
     . explode('?', $_SERVER['REQUEST_URI'], 2)[0];

...returns, for example:

http://example.com/directory/file.php

---

php.com Documentation:

• $_SERVER — Server and execution environment information
• explode — Split a string by a string
• parse_url — Parse a URL and return its components (possibly a better solution)

Answer 6

You can use $_GET for url params, or $_POST for post params, but the $_REQUEST contains the parameters from $_GET $_POST and $_COOKIE, if you want to hide the URI parameter from the user you can convert it to a session variable like so:

<?php

session_start();
if (isset($_REQUEST['param']) && !isset($_SESSION['param'])) {

    // Store all parameters received
    $_SESSION['param'] = $_REQUEST['param'];

    // Redirect without URI parameters
    header('Location: /file.php');
    exit;
}
?>
<html>
<body>
<?php
  echo $_SESSION['param'];
?>
</body>
</html>

EDIT

use $_SERVER['PHP_SELF'] to get the current file name or $_SERVER['REQUEST_URI'] to get the requested URI