getcwd()
(PHP 4, PHP 5, PHP 7)
取得当前工作目录
说明
getcwd(void): string
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 #1getcwd()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
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.
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.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_DirStack, getcwd() );
chdir( $dir );
}
function popd( )
{
global $g_DirStack;
$dir = array_pop( $g_DirStack );
assert( $dir !== null );
chdir( $dir );
}
?>
This allows you to change the current directory with pushd, then use popd to "undo" the directory change when you're done.It appears there is a change in functionality in PHP5 from PHP4 when using the CLI tool. Here is the example: - cd /tmp cat > foo.php << END <?php print getcwd() . "\n"; ?> END cd / php -q /tmp/foo.php PHP4 returns /tmp PHP5 returns / Something to be aware of.
This function is often used in conjuction with basename(), i.e. http://www.php.net/manual/en/function.basename.php
When running PHP on the command line, if you want to include another file which is in the same directory as the main script, doing just
<?php
include './otherfile.php';
?>
might not work, if you run your script like this:
/$ /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 $dir;
}
?>
So you can use it like this:
<?php
include get_file_dir() . '/otherfile.php';
// or even..
chdir(get_file_dir());
include './otherfile.php';
?>
Spent some time thinking this one out, maybe it helps someone :)This is a function to convert a path which looks something like this:
/home/www/somefolder/../someotherfolder/../
To a proper directory path:
<?php
function simplify_path($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" state. It works in *NIX and WINDOWS alike
?>
"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.
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
Some server's has security options to block the getcwd() Alternate option: str_replace($_SERVER['SCRIPT_NAME'],'', $_SERVER['SCRIPT_FILENAME']);
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 , realpath( getcwd() ) ); // how to switch symlink-free to the folder the current file resides in: chdir( dirname ( realpath ( __FILE__ ) ) ); // how to store the former folder in a variable: $my_folder = dirname( realpath( __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 ); ?>
watch out: working directory, and thus: getcwd () is "/" while being into a register'ed shutdown function!!!
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.
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
$cwd = exec('pwd');
?>
In response to myself: that function will not work for cases like:
/usr/bin$: /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 :)Be aware when calling getcwd() in directories consisting of symlinks.
getcwd() is the equivalent of shell command "pwd -P" which resolves symlinks.
The shell command "pwd" is the equivalent of "pwd -L" which uses PWD from the environment without resolving symlinks. This is also the equivalent of calling getenv('PWD').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.
getcwd() appears to call the equivalent of PHP's realpath() on the path. It never returns symlinks, but always the actual directory names in the path to the current working directory.
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.