Get only filename from url in php without any variable values which exist in the url

Question

I want to get filename without any $_GET variable values from a URL in php?

My URL is http://learner.com/learningphp.php?lid=1348

I

Answer 1

Try the following code:

For PHP 5.4.0 and above:

$filename = basename(parse_url('http://learner.com/learningphp.php?lid=1348')['path']);

For PHP Version < 5.4.0

$parsed = parse_url('http://learner.com/learningphp.php?lid=1348');
$filename = basename($parsed['path']);

Answer 2

Use this function:

function getScriptName()
{
    $filename = baseName($_SERVER['REQUEST_URI']);
    $ipos = strpos($filename, "?");
    if ( !($ipos === false) )   $filename = substr($filename, 0, $ipos);
    return $filename;
}

Answer 3

Is better to use parse_url to retrieve only the path, and then getting only the filename with the basename. This way we also avoid query parameters.

<?php

// url to inspect
$url = 'http://www.example.com/image.jpg?q=6574&t=987';

// parsed path
$path = parse_url($url, PHP_URL_PATH);

// extracted basename
echo basename($path);

?>

Is somewhat similar to Sultan answer excepting that I'm using component parse_url parameter, to obtain only the path.

Answer 4

Your URL:

$url = 'http://learner.com/learningphp.php?lid=1348';
$file_name = basename(parse_url($url, PHP_URL_PATH));
echo $file_name;

output: learningphp.php

Answer 5

May be i am late

$e = explode("?",basename($_SERVER['REQUEST_URI']));
$filename = $e[0];

Answer 6

Use parse_url() as Pekka said:

<?php
$url = 'http://www.example.com/search.php?arg1=arg2';

$parts = parse_url($url);

$str = $parts['scheme'].'://'.$parts['host'].$parts['path'];

echo $str;
?>

http://codepad.org/NBBf4yTB

In this example the optional username and password aren't output!