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

问题

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? Please help.

I used basename but it gives all the variable values also- learntolearn.php?lid=1348 which are in the url.

回答1:

This should work:

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

But beware of any malicious parts in your URL.

回答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

回答3:

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!

回答4:

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.

回答5:

You can use,

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

echo $directoryURI;

回答6:

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'

回答7:

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']);

回答8:

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

回答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

回答10:

Use this function:

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

来源：https://stackoverflow.com/questions/7852296/get-only-filename-from-url-in-php-without-any-variable-values-which-exist-in-the

标签

php

url

get

filenames