Get only filename from url in php without any variable values which exist in the url
PhpUrlGetFilenamesPhp Problem Overview
I want to get filename without any $_GET
variable values from a URL in php?
My URL is http://learner.com/learningphp.php?lid=1348
I only want to retrieve the learningphp.php
from the URL?
How to do this?
I used basename()
function but it gives all the variable values also: learntolearn.php?lid=1348
which are in the URL.
Php Solutions
Solution 1 - Php
This should work:
echo basename($_SERVER['REQUEST_URI'], '?' . $_SERVER['QUERY_STRING']);
But beware of any malicious parts in your URL.
Solution 2 - Php
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
Solution 3 - Php
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.
Solution 4 - Php
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;
?>
In this example the optional username and password aren't output!
Solution 5 - Php
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
Solution 6 - Php
You can use,
$directoryURI =basename($_SERVER['SCRIPT_NAME']);
echo $directoryURI;
Solution 7 - Php
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'
Solution 8 - Php
$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
Solution 9 - Php
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']);
Solution 10 - Php
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.
Solution 11 - Php
$url = "learner.com/learningphp.php?lid=1348";
$l = parse_url($url);
print_r(stristr($l['path'], "/"));
Solution 12 - Php
Use this function:
function getScriptName()
{
$filename = baseName($_SERVER['REQUEST_URI']);
$ipos = strpos($filename, "?");
if ( !($ipos === false) ) $filename = substr($filename, 0, $ipos);
return $filename;
}
Solution 13 - Php
May be i am late
$e = explode("?",basename($_SERVER['REQUEST_URI']));
$filename = $e[0];