Directory 函数
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getcwd

(PHP 4, PHP 5)

getcwd取得当前工作目录

说明

string getcwd ( void )

取得当前工作目录。

返回值

成功则返回当前工作目录,失败返回 FALSE

在某些 Unix 的变种下,如果任何父目录没有设定可读或搜索模式,即使当前目录设定了, getcwd() 还是会返回 FALSE。有关模式与权限的更多信息见 chmod()

范例

Example #1 getcwd() 例子

<?php

// current directory
echo getcwd() . "\n";

chdir('cvs');

// current directory
echo getcwd() . "\n";

?>

以上例程的输出类似于:

/home/didou
/home/didou/cvs

参见


Directory 函数
在线手册:中文  英文

用户评论:

Anonymous (2010-07-14 04:04:41)

As you could read in
http://www.php.net/manual/en/features.commandline.differences.php
the CLI SAPI does - contrary to other SAPIs - NOT automatically change the current working directory to the one the started script resides in.

A very simple workaround to regain the behaviour you're used to from your "ordinary" webpage scripting is to include something like that at the beginning of your script:

<?php
  chdir
dirname __FILE__ ) );
?>

But because this is about reading or "finding" pathes, you might appreciate it if I share some more sophisticated tricks I frequently use in CLI scripts ...

<?php
// Note: all pathes stored in subsequent Variables end up with a DIRECTORY_SEPARATOR

// how to store the working directory "from where" the script was called:
$initial_cwd preg_replace'~(\w)$~' '$1' DIRECTORY_SEPARATOR realpathgetcwd() ) );

// how to switch symlink-free to the folder the current file resides in:
chdirdirname realpath __FILE__ ) ) );

// how to store the former folder in a variable:
$my_folder dirnamerealpath__FILE__ ) ) . DIRECTORY_SEPARATOR;

// how to get a path one folder up if $my_folder ends with \class\ or /class/ :
$my_parent_folder preg_replace'~[/\\\\]class[/\\\\]$~' DIRECTORY_SEPARATOR $my_folder );

// how to get a path one folder up in any case :
$my_parent_folder preg_replace'~[/\\\\][^/\\\\]*[/\\\\]$~' DIRECTORY_SEPARATOR $my_folder );

// how to make an array of OS-style-pathes from an array of unix-style-pathes
// (handy if you use config-files or so):
foreach( $unix_style_pathes as $unix_style_path )
    
$os_independent_path[] = str_replace'/' DIRECTORY_SEPARATOR $unix_style_path );

?>

bvidinli at gmail dot com (2009-01-21 01:13:24)

if you link your php to /bin/linkedphp and your php is at for ex /home/actual.php
when you run linkedphp in somewhere in your filesystem,
getcwd returns /bin instead of working dir,
solution: use dirname(__FILENAME__) instead

ash at ashmckenzie dot org (2008-11-18 17:37:37)

It appears there is a change in functionality in PHP5 from PHP4 when using the CLI tool.  Here is the example: -

cd /tmp

cat > foo.php << END
<?php
    
print getcwd() . "\n";
?>
END

cd /

php -q /tmp/foo.php

PHP4 returns /tmp
PHP5 returns /

Something to be aware of.

Wolfgang M. Pauli (2008-05-04 19:44:39)

If you try to use getcwd() in a directory that is a symbolic link, getcwd() gives you the target of that link (similarly when parent etc. is symbolic link). There might be a better solution, but this worked for me (linux):
<? php
$cwd = exec('pwd');
?>

znupi69NOSPAMHERE at gmail dot com (2008-03-03 10:39:09)

In response to myself: that function will not work for cases like:
/usr/bin$: /home/johndoe/Work/script.php
So here's a better and simpler way (I think this one works for all cases)
<?php
function get_file_dir() {
    global 
$argv;
    return 
realpath($argv[0]);
}
?>
Knock yourself out :)

znupi69NOSPAMHERE at gmail dot com (2008-03-02 07:01:48)

When running PHP on the command line, if you want to include another file which is in the same directory as the main script, doing just
<?php
include './otherfile.php';
?>
might not work, if you run your script like this:
/$ /path/to/script.php
because the current working dir will be set to '/', and the file '/otherfile.php' does not exist, because it is in '/path/to/otherfile.php'.
So, to get the directory in which the script resides, you can use this function:
<?php
function get_file_dir() {
    global 
$argv;
    
$dir dirname(getcwd() . '/' $argv[0]);
    
$curDir getcwd();
    
chdir($dir);
    
$dir getcwd();
    
chdir($curDir);
    return 
$dir;
}
?>
So you can use it like this:
<?php
include get_file_dir() . '/otherfile.php';
// or even..
chdir(get_file_dir());
include 
'./otherfile.php';
?>
Spent some time thinking this one out, maybe it helps someone :)

leonbrussels at gmail dot com (2007-12-17 08:10:26)

This is a function to convert a path which looks something like this:

/home/www/somefolder/../someotherfolder/../

To a proper directory path:

<?php

function simplify_path($path) {

//saves our current working directory to a variable
$oldcwd getcwd();
//changes the directory to the one to convert
//$path is the directory to convert (clean up), handed over to the //function as a string

chdir($path);
return 
gstr_replace('\\''/'getcwd());

//change the cwd back to the old value to not interfere with the script
chdir($oldcwd);

}

This function is really useful if you want to compare two filepaths which are not necesarily in a "cleaned up" stateIt works in *NIX and WINDOWS alike

?>

ab5602 at wayne dot edu (2007-09-30 10:33:15)

If getcwd() returns nothing for you under Solaris with an NFS mounted subdirectory, you are running into an OS bug that is supposedly fixed in recent versions of Solaris 10. This same OS bug effects the include() and require() functions as well.

troy dot cregger at gmail dot com (2007-01-26 12:10:31)

Take care if you use getcwd() in file that you'll need to include (using include, require, or *_once) in a script located outside of the same directory tree. 

example: 
<?php
//in /var/www/main_document_root/include/MySQL.inc.php
if (strpos(getcwd(),'main_')>0) {
  
//code to set up main DB connection
}
?>

<?php
//in home/cron_user/maintenance_scripts/some_maintenance_script.php
require_once ('/var/www/main_document_root/include/MySQL.inc.php');
?>

In the above example, the database connection will not be made because the call to getcwd() returns the path relative to the calling script ( /home/cron_user/maintenance_scripts ) NOT relative to the file where the getcwd() function is called.

mark dot phpnetspam at mhudson dot net (2006-11-03 13:42:40)

This function is often used in conjuction with basename(), i.e.
http://www.php.net/manual/en/function.basename.php

hodgman at ali dot com dot au (2006-09-06 19:58:33)

I use this code to replicate the pushd and popd DOS commands in PHP:

<?php
$g_DirStack 
= array();
function 
pushd$dir )
{
    global 
$g_DirStack;
    
array_push$g_DirStackgetcwd() );
    
chdir$dir );
}
function 
popd( )
{
    global 
$g_DirStack;
    
$dir array_pop$g_DirStack );
    
assert$dir !== null );
    
chdir$dir );
}
?>

This allows you to change the current directory with pushd, then use popd to "undo" the directory change when you're done.

emailfire at gmail dot com (2005-12-08 06:57:03)

To get the username of the account:

<?php
$dir 
getcwd();
$part explode('/'$dir);
$username $part[1];
?>

If current directory is '/home/mike/public_html/' it would return mike.

memandeemail at gmail dot com (2005-12-07 03:48:14)

Some server's has security options to block the getcwd()
Alternate option:
str_replace($_SERVER['SCRIPT_NAME'],'', $_SERVER['SCRIPT_FILENAME']);

marcus at synchromedia dot co dot uk (2005-05-04 08:41:26)

"On some Unix variants, getcwd() will return FALSE if any one of the parent directories does not have the readable or search mode set, even if the current directory does."
Just so you know, MacOS X is one of these variants (at least 10.4 is for me). You can make it work by applying 'chmod a+rx' to all folders from your site folder upwards.

vermicin at antispam dot gmail dot com (2005-01-27 03:48:50)

If your PHP cli binary is built as a cgi binary (check with php_sapi_name), the cwd functions differently than you might expect.
say you have a script /usr/local/bin/purge
you are in /home/username
php CLI: getcwd() gives you /home/username
php CGI: getcwd() gives you /usr/local/bin
This can trip you up if you're writing command line scripts with php. You can override the CGI behavior by adding -C to the php call:
#!/usr/local/bin/php -Cq
and then getcwd() behaves as it does in the CLI-compiled version.

manux at manux dot org (2004-06-21 21:44:21)

watch out:
working directory, and thus:
getcwd ()
is "/" while being into a register'ed shutdown function!!!

dave at corecomm dot us (2003-12-10 10:14:17)

getcwd() returns the path of the "main" script referenced in the URL.
dirname(__FILE__) will return the path of the script currently executing.
I had written a script that required several class definition scripts from the same directory. It retrieved them based on filename matches and used getcwd to figure out where they were.
Didn't work so well when I needed to call that first script from a new file in a different directory.

fvu at wanadoo dot nl (2003-11-23 17:50:07)

Make sure to lowercase the result before comparing Windows paths, because this function returns ambiguous results on Windows.  From within Apache, the returned path is lowercase, while from the command line interface (CLI) the returned path uses the 'real' Windows pathname.  For example, running the 'print getcwd();' command from 'C:\Program Files' returns either

   c:\program files   (Apache)
   C:\Program Files   (CLI)

When the directory is specified using chdir(), getcwd() uses the exact chdir argument.  For example:

   <?php chdir('C:\\PrOgRaM fIlEs'); print getcwd(); ?>

outputs:

   C:\PrOgRaM fIlEs   (Apache & CLI)

The following code can be used to return a unambiguous lowercased cwd when running on Windows:

   <?php $sCwd = (substr(PHP_OS03) == 'WIN') ? strtolower(getcwd()) : getcwd(); ?>

raja at rajashahed dot com (2003-11-06 10:21:42)

This is current working directory. X example, your document root is c:\Inetpub\www\htdocs. When You need to know what is your doc_root; /* Like ask yourself your name ;)*/
$_cur_dir = getcwd();
echo "My doc_root is $_cur_dir ";
// it prints out : My doc_root is c:\Inetpub\www\htdocs
/* Usually you need it after using chdir() to know what is
running in current directory */
Regards
Raja

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