CODEWITHSUNDEEP
phpurlgetfilenames
sqlchild
sqlchild

Reputation: 9084

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

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.

Upvotes: 71

Views: 162282

Answers (13)

daprezjer
daprezjer

Reputation: 903

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.

Upvotes: 2

Rachit Vohera
Rachit Vohera

Reputation: 745

May be i am late

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

Upvotes: 1

Udhav Sarvaiya
Udhav Sarvaiya

Reputation: 10111

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

Upvotes: 24

foxtrot
foxtrot

Reputation: 1096

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.

Upvotes: 38

Khadka Pushpendra
Khadka Pushpendra

Reputation: 999

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

Upvotes: 42

Konstantin XFlash Stratigenas
Konstantin XFlash Stratigenas

Reputation: 607

Use this function:

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

Upvotes: 1

Sultan
Sultan

Reputation: 647

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']);

Upvotes: 5

SwR
SwR

Reputation: 632

You can use,

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

echo $directoryURI;

Upvotes: 10

James
James

Reputation: 41

$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

Upvotes: 4

ᴄʀᴏᴢᴇᴛ
ᴄʀᴏᴢᴇᴛ

Reputation: 2999

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'

Upvotes: 6

Mob
Mob

Reputation: 11106

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

Upvotes: 1

str
str

Reputation: 45029

This should work:

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

But beware of any malicious parts in your URL.

Upvotes: 84

ComFreek
ComFreek

Reputation: 29444

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!

Upvotes: 22

Related Questions