TO stephane dot fidanza at gmail dot com
i don't think file with name you given can exists...
basename
(PHP 4, PHP 5)
basename — パス中のファイル名の部分を返す
説明
string basename ( string $path [, string $suffix] )この関数は、ファイルへのパスを有する文字列を引数とし、 ファイルのベース名を返します。ファイル名が、 suffix で終了する場合、 この部分もカットされます。
Windows では、スラッシュ(/) とバックスラッシュ (\) の両方がディレクトリ区切り文字として使われます。 その他の環境ではスラッシュ(/)になります。
例 605. basename() の例
<?php
$path = "/home/httpd/html/index.php";
$file = basename($path); // $file は "index.php" に設定される
$file = basename($path,".php"); // $file は "index" に設定される
?>
注意: パラメータ suffixは、 PHP 4.1.0 で追加されました。
dirname() も参照ください。
basename
maksymus007 at wp dot pl
22-Jul-2007 04:08
22-Jul-2007 04:08
stephane dot fidanza at gmail dot com
11-Apr-2007 08:33
11-Apr-2007 08:33
Support of the $suffix parameter has changed between PHP4 and PHP5:
in PHP4, $suffix is removed first, and then the core basename is applied.
conversely, in PHP5, $suffix is removed AFTER applying core basename.
Example:
<?php
$file = "path/to/file.xml#xpointer(/Texture)";
echo basename($file, ".xml#xpointer(/Texture)");
?>
Result in PHP4: file
Result in PHP5: Texture)
phdwight at yahoo dot com
22-Mar-2007 11:41
22-Mar-2007 11:41
Pulled my hair out for this.
Just like most functions, this does not play well with Japanese characters.
When you call basename with some Japanese characters, it will return nothing (truncates it) .
thoughts at highermind dot org
30-Jan-2007 10:45
30-Jan-2007 10:45
Basename without query string:
<?php
$filename = array_shift(explode('?', basename($url_path)));
?>
jonmsawyer at gmail dot com
10-Jan-2007 05:23
10-Jan-2007 05:23
@antrik at users dot sf dot net
> 15-Nov-2004 10:40
> 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!
There is a reason behind this -- and it has nothing to do with being a feature. PHP was heavily modeled off of the POSIX model. Many of the same functions you see in PHP are also in C, C++, and Java. These languages are modeled on POSIX as well.
The directory '/bar/foo/', when passed into the function basename(), will output 'foo' because *everything*, including directories, in the POSIX model, is a _file_. Most unix platforms, and all Windows platforms are (some Linux distributions are not) fully compliant to the POSIX model.
For example, the device file that contains information about your harddisk, in Linux, is probably stored in the _file_ /dev/hda.
Another example is that when you want to list information about your CPU or Memory using the Linux kernel, you might read the _file_ /proc/cpu/info.
Directories are no exception. Directories are no more different than your regular text file -- other than the fact that they describe a _file_-list of all files under it, and where the OS can access them. This means that even directories treat other directories as files.
The reason why we are made to think that directories are not files is because the kernel (the OS) simply treats these culprits differently. Your OS is lying to you! When you try to open up c:\windows in Notepad, you simply get a runaround because the Windows operating system knows it is a directory and knows how to treat it -- and knowing this it will not let you open it up for editing. For if you did that, you would probably lose the data in that directory. If you are familiar with C programming, you will know that if you lose information about a pointer to an object, the object gets lost in memory. The same would happen if you modified a directory in the wrong way. This is why the operating system protects its directories with the upmost care. (Some do anyway, hehe)
So when doing any kind of programming in PHP, C/++, Java, Ada, Perl, Python, Ruby, FORTRAN, and yes, even RPG IV (for all of you AS/400 folks out there working on the IFS), you must treat directories as files well.
This is why 'foo' is returned. For more information on POSIX, see http://en.wikipedia.org/wiki/POSIX
I hope this helps. Cheers.
thovan at gmx dot net
03-Jan-2007 12:24
03-Jan-2007 12:24
After reading all of the earlier comments, I made my own function file_basename():
<?php
function file_basename($file= null) {
if($file=== null || strlen($file)<= 0) {
return null;
}
$file= explode('?', $file);
$file= explode('/', $file[0]);
$basename= $file[count($file)-1];
return $basename;
}
?>
19-Sep-2006 03:28
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 09:19
18-Feb-2006 09:19
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 04:55
18-Feb-2006 04:55
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 09:18
05-Jan-2006 09:18
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 09:57
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 05:44
02-Nov-2005 05:44
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 04:47
25-Oct-2005 04:47
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 04:38
23-Sep-2005 04:38
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 10:53
05-Sep-2005 10:53
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 09:47
23-Aug-2005 09:47
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 01:28
15-Jul-2005 01:28
Note that in my example below, I used the stripslashes function on the target string first because I was dealing with the POST array $_FILES. When creating this array, PHP will add slashes to any slashes it finds in the string, so these must be stripped out first before processing the file path. Then again, the only reason I can think of that basename() would fail is when dealing with Windows paths on a *nix server -- and the file upload via POST is the only situation I can think of that would require this. Obviously, if you are not dealing with these additional slashes, invoking stripslashes() first would remove the very separators you need extract the file name from the full path.
amitabh at NOSPAM dot saysnetsoft dot com
14-Jul-2005 05:55
14-Jul-2005 05:55
The previous example posted by "pvollma" didn't work out for me, so I modified it slightly:
<?php
function GetFileName($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 03:43
14-Jul-2005 03:43
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 11:18
30-Jan-2005 11:18
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 09:24
30-Jan-2005 09:24
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 03:40
16-Nov-2004 03:40
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($path) ?> Watch out!
osanim at cidlisuis dot org
18-Apr-2004 03:12
18-Apr-2004 03:12
If you want know the real directory of the include file, you have to writte:
<?php
dirname(__FILE__)
?>
KOmaSHOOTER at gmx dot de
28-Nov-2003 07:33
28-Nov-2003 07:33
Exmaple for exploding ;) the filename to an array
<?php
echo(basename ($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
09-Nov-2003 12:52
09-Nov-2003 12:52
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 02:22
16-Oct-2003 02:22
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 06:53
02-Apr-2003 06:53
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>";
?>