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 only want to retrieve the learningphp.php from the URL?

How to do this?

I used basename() function but it gives all the variable values also: learntolearn.php?lid=1348 which are in the URL.

Answer 1

This should work:

echo basename($_SERVER['REQUEST_URI'], '?' . $_SERVER['QUERY_STRING']);

But beware of any malicious parts in your URL.

Answer 2

Following steps shows total information about how to get file, file with extension, file without extension. This technique is very helpful for me. Hope it will be helpful to you too.

  $url = 'https://www.google.com/images/branding/googlelogo/2x/googlelogo_color_120x44dp.png';
        $file = file_get_contents($url); // to get file
        $name = basename($url); // to get file name
        $ext = pathinfo($url, PATHINFO_EXTENSION); // to get extension
        $name2 =pathinfo($url, PATHINFO_FILENAME); //file name without extension

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

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!

Answer 6

You can use,

$directoryURI =basename($_SERVER['SCRIPT_NAME']);

echo $directoryURI;

Answer 7

An other way to get only the filename without querystring is by using parse_url and basename functions :

$parts = parse_url("http://example.com/foo/bar/baz/file.php?a=b&c=d");
$filename = basename($parts["path"]); // this will return 'file.php'

Answer 8

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 9

$filename = pathinfo( parse_url( $url, PHP_URL_PATH ), PATHINFO_FILENAME ); 

Use parse_url to extract the path from the URL, then pathinfo returns the filename from the path

Answer 10

The answer there assumes you know that the URL is coming from a request, which it may very well not be. The generalized answer would be something like:

$basenameWithoutParameters = explode('?', pathinfo($yourURL, PATHINFO_BASENAME))[0];

Here it just takes the base path, and splits out and ignores anything ? and after.

Answer 11

Use this function:

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

Answer 12

$url = "learner.com/learningphp.php?lid=1348";
$l = parse_url($url);
print_r(stristr($l['path'], "/"));

Answer 13

May be i am late

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