basename

add a note User Contributed Notes

19-Sep-2006 02:28


lazy lester is just confirming what icewind said.

And yes it is correct! unlike what the following comment after icewind says, as that example is the same with the line order reversed! as poniestail at gmail dot com says.



But poniestail at gmail dot com missed the point that if the url is coming from a log file it will not have its value in $_SERVER["QUERY_STRING"] or $_SERVER["SCRIPT_NAME"] but in a LOG FILE or a DATABASE

lazy lester
18-Feb-2006 08:19


If your path has a query string appended, and if the query string contains a "/" character, then the suggestions for extracting the filename offered below don't work. 



For instance if the path is like this:

http://www.ex.com/getdat.php?dep=n/a&title=boss



Then both the php basename() function, and also 

the $_SERVER[QUERY_STRING] variables get confused.



In such a case, use:



<php

$path_with_query="http://www.ex.com/getdat.php?dep=n/a&title=boss";

$path=explode("?",$path_with_query);

$filename=basename($path[0]);

$query=$path[1];

?>

support at rebootconcepts dot com
18-Feb-2006 03:55


works on windows and linux, faster/easier than amitabh's...



<?php

$basename = preg_replace( '/^.+[\\\\\\/]/', '', $filename );



// Optional; change any non letter, hyphen, or period to an underscore.

$sterile_filename = preg_replace( "/[^\w\.-]+/", "_", $basename );

?>

poniestail at gmail dot com
05-Jan-2006 08:18


examples from "icewind" and "basname" seem highly overdone... not to mention example from "basename" is exactly the same as one from "icewind"...



possibly a more logical approach?

<?

   //possible URL = http://domain.com/path/to/file.php?var=foo

   $filename = substr( $_SERVER["SCRIPT_NAME"], 1 ); //substr( ) used for optional removal of initial "/"

   $query = $_SERVER["QUERY_STRING"];

?>



to see the entire $_SERVER variable try this:

<?

   echo "<pre>

      ".print_r( $_SERVER, true )."

      </pre>

   ";

?>

15-Nov-2005 08:57


icewinds exmaple wouldn't work, the query part would contain the second char of the filename, not the query part of the url.

<?

$file = "path/file.php?var=foo";

$file = explode("?", basename($file));

$query = $file[1];

$file = $file[0];

?>



That works better.

icewind
02-Nov-2005 04:44


Because of filename() gets "file.php?var=foo", i use explode in addition to basename like here:



$file = "path/file.php?var=foo";

$file = explode("?", basename($file));

$file = $file[0];

$query = $file[1];



Now $file only contains "file.php" and $query contains the query-string (in this case "var=foo").

www.turigeza.com
25-Oct-2005 03:47


simple but not said in the above examples



echo basename('somewhere.com/filename.php?id=2', '.php');

will output

filename.php?id=2



which is not the filename in case you expect!

crash at subsection dot org dot uk
23-Sep-2005 03:38


A simple way to return the current directory:

$cur_dir = basename(dirname($_SERVER[PHP_SELF]))



since basename always treats a path as a path to a file, e.g.



/var/www/site/foo/ indicates /var/www/site as the path to file 

foo

tomboshoven at gmail dot com
05-Sep-2005 09:53


basename() also works with urls, eg:



basename('http://www.google.com/intl/en/images/logo.gif');



will return 'logo.gif'.

b_r_i_a__n at locallinux dot com
23-Aug-2005 08:47


I was looking for a way to get only the filename whether or not I had received the full path to it from the user.  I came up with a much simpler (and probably more robust) method by using the power of basename in reverse:



$infile = "/usr/bin/php";

$filename = stristr ($infile,basename ($infile));



This even works on those _wacky_ filenames like "/usr/lib/libnetsnmp.so.5.0.9" which are not factored when exploding the full path and taking out only the last segment after "."

pvollma at pcvsoftware dot net
15-Jul-2005 12:28


Note that in my example below, I used the stripslashes function on the target string first because I was dealing with the POST array $_FILES. When creating this array, PHP will add slashes to any slashes it finds in the string, so these must be stripped out first before processing the file path. Then again, the only reason I can think of that basename() would fail is when dealing with Windows paths on a *nix server -- and the file upload via POST is the only situation I can think of that would require this. Obviously, if you are not dealing with these additional slashes, invoking stripslashes() first would remove the very separators you need extract the file name from the full path.

amitabh at NOSPAM dot saysnetsoft dot com
14-Jul-2005 04:55


The previous example posted by "pvollma" didn't work out for me, so I modified it slightly:

<?php

function GetFileName($file_name)

{

        $newfile = basename($file_name);

        if (strpos($newfile,'\\') !== false)

        {

                $tmp = preg_split("[\\\]",$newfile);

                $newfile = $tmp[count($tmp) - 1];

                return($newfile);

        }

        else

        {

                return($file_name);

        }

}

?>

pvollma at pcvsoftware dot net
14-Jul-2005 02:43


There is a real problem when using this function on *nix servers, since it does not handle Windows paths (using the \ as a separator). Why would this be an issue on *nix servers? What if you need to handle file uploads from MS IE? In fact, the manual section "Handling file uploads" uses basename() in an example, but this will NOT extract the file name from a Windows path such as C:\My Documents\My Name\filename.ext. After much frustrated coding, here is how I handled it (might not be the best, but it works):



<?php

$filen = stripslashes($_FILES['userfile']['name']);

$newfile = basename($filen);

if (strpos($newfile,'\\') !== false) {

  $tmp = preg_split("[\\\]",$newfile);

  $newfile = $tmp[count($tmp) - 1];

}

?>



$newfile will now contain only the file name and extension, even if the POSTed file name included a full Windows path.

KOmaSHOOTER at gmx dot de
30-Jan-2005 10:18


if you want the name of the parent directory

<?php

$_parenDir_path = join(array_slice(split( "/" ,dirname($_SERVER['PHP_SELF'])),0,-1),"/").'/'; // returns the full path to the parent dir

$_parenDir =  basename ($_parenDir_path,"/"); // returns only the name of the parent dir

// or 

$_parenDir2 = array_pop(array_slice(split( "/" ,dirname($_SERVER['PHP_SELF'])),0,-1)); // returns also only the name of the parent dir

echo('$_parenDir_path  = '.$_parenDir_path.'<br>');

echo('$_parenDir  = '.$_parenDir.'<br>');

echo('$_parenDir2  = '.$_parenDir2.'<br>');

?>

KOmaSHOOTER at gmx dot de
30-Jan-2005 08:24


If you want the current path where youre file is and not the full path then use this :)



<?php

echo('dir = '.basename (dirname($_SERVER['PHP_SELF']),"/"));    

// retuns the name of current used directory

?>



Example: 



www dir: domain.com/temp/2005/january/t1.php



<?php

echo('dirname <br>'.dirname($_SERVER['PHP_SELF']).'<br><br>');    

// returns: /temp/2005/january

?>



<?php

echo('file = '.basename ($PHP_SELF,".php"));    

// returns: t1

?>



if you combine these two you get this

<?php 

echo('dir = '.basename (dirname($_SERVER['PHP_SELF']),"/"));    

// returns: january

?>



And for the full path use this

<?php 

echo(' PHP_SELF <br>'.$_SERVER['PHP_SELF'].'<br><br>');

// returns: /temp/2005/january/t1.php    

?>

antrik at users dot sf dot net
16-Nov-2004 02:40


When using basename() on a path to a directory ('/bar/foo/'), the last path component ('foo') is returned, instead of the empty string one would expect. (Both PHP 4.1.2 and 4.3.8 on GNU/Linux.) No idea whether this is considered a bug or a feature -- I found it extremely annoying. Had to work around using: <?php $file=substr($path, -1)=='/'?'':basename($path) ?> Watch out!

osanim at cidlisuis dot org
18-Apr-2004 02:12


If you want know the real directory of the include file, you have to writte:





<?php


dirname(__FILE__)


?>

KOmaSHOOTER at gmx dot de
28-Nov-2003 06:33


Exmaple for exploding ;) the filename to an array 



<?php

echo(basename ($PHP_SELF)."<br>");  // returnes filename.php

$file = basename ($PHP_SELF);

$file = explode(".",$file);

print_r($file);    // returnes Array ( [0] => filename [1] => php ) 

echo("<br>");

$filename = basename(strval($file[0]),$file[1]); 

echo($filename."<br>");  // returnes  filename

echo(basename ($PHP_SELF,".php")."<br>");  // returnes  filename

echo("<br>");

echo("<br>");

//show_source(basename ($PHP_SELF,".php").".php")

show_source($file[0].".".$file[1])

?>

giovanni at giacobbi dot net
08-Nov-2003 11:52


No comments here seems to take care about UNIX system files, which typically start with a dot, but they are not "extensions-only".

The following function should work with every file path. If not, please let me know at my email address.



<?php



function remove_ext($str) {

  $noext = preg_replace('/(.+)\..*$/', '$1', $str);

  print "input: $str\n";

  print "output: $noext\n\n";

}



remove_ext("/home/joh.nny/test.php");

remove_ext("home/johnny/test.php");

remove_ext("weirdfile.");

remove_ext(".hiddenfile");

remove_ext("../johnny.conf");

daijoubu_NOSP at M_videotron dot ca
16-Oct-2003 01:22


An faster alternative to:



<?php

array_pop(explode('.', $fpath));

?>



would be:



<?php

substr($fpath, strrpos($fpath, '.')); // returns the dot

?>



If you don't want the dot, simply adds 1 to the position



<?php

substr($fpath, strrpos($fpath, '.') + 1); // returns the ext only

?>

Richard at Lyders dot Net
02-Apr-2003 05:53


you can also make use of the basename() function's second parameter:



<?PHP

$fpath = "/blah/file.name.has.lots.of.dots.ext";

$fext  = array_pop(explode('.', $fpath));

$fname = basename($fpath, '.'.$fext);



print "fpath: $fpath\n<br>";

print "fext: $fext\n<br>";

print "fname: $fname\n<br>";

?>

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basename

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