is_file
To find out whether a file is hardlinked to another filename, check the number of links of the stat() output. If it is >1 there is another filename for that file.
To find out whether two filenames are pointing to the same file, check the inode number of those 2 filenames. If it is equal, the 2 filenames are hardlinked together.
is_link
(PHP 4, PHP 5)
is_link — 判断给定文件名是否为一个符号连接
说明
bool is_link
( string $filename
)
如果文件存在并且是一个符号连接则返回 TRUE。
Note: 本函数的结果会被缓存。更多信息参见 clearstatcache()。
Tip
自 PHP 5.0.0 起本函数也可被某些 URL wrapper 使用。参考支持的协议／封装协议列表 来看哪些 wrapper 支持 stat() 系列函数的功能。
参见 is_dir()，is_file() 和 readlink()。
add a note
User Contributed Notesis_link
mbirth at webwriters dot de
20-May-2008 11:24
20-May-2008 11:24
neverpanic->gmail[com]
30-Dec-2006 06:50
30-Dec-2006 06:50
For me (Debian Sarge VPS) is_link returns true even for directories if you don't add a trailing slash to the filename.
<?php
if ($dir{strlen($dir)-1} == '/') $dir = substr($dir, 0, -1);
is_link($dir);
?>
This works for me. It can't detect a symlink somewhere in a complete path, though (i.e. is_link(/www/somedir/file.php) will return false, just as is_link(/www/) would)
brendy at gmail dot com
06-May-2006 10:22
06-May-2006 10:22
On Mac OSX, to see if a file is a FInder alias:
<?PHP
if( getFinderAlias( $someFile , $target ) ) {
echo $target;
}
else {
echo "File is not an alias";
}
function getFinderAlias( $filename , &$target ) {
$getAliasTarget = <<< HEREDOC
-- BEGIN APPLESCRIPT --
set checkFileStr to "{$filename}"
set checkFile to checkFileStr as POSIX file
try
tell application "Finder"
if original item of file checkFile exists then
set targetFile to (original item of file checkFile) as alias
set posTargetFile to POSIX path of targetFile as text
get posTargetFile
end if
end tell
end try
-- END APPLESCRIPT --
HEREDOC;
$runText = "osascript << EOS\n{$getAliasTarget}\nEOS\n";
$target = trim( shell_exec( $runText ) );
return ( $target == "" ? false : true );
}
?>
jr at cnb dot uam dot es
31-May-2005 02:31
31-May-2005 02:31
Why don't you just try
is_dir("$pathname/.")
instead?
If $pathname is a directory, $pathname/. is itself and is a directory too.
If $pathname is a link to a directory, then $pathname/. is the actual directory pointed at and is a directory as well.
If $pathname is a link to a non-directory, then $pathname/. does not exist and returns FALSE, as it should.
A lot easier, more readable and intuitive.
andudi at gmx dot ch
03-Jun-2002 01:44
03-Jun-2002 01:44
On my SuSE 7.2 is_link does not work on directories, but to find out, if a dir is a link, I use now this:
$linkdir = $path.$linkdirname;
if (realpath($linkdir) != realpath($path)."/".$linkdirname):
//$linkdir is a symbolic linked dir!
...
and this works fine :-)
Andreas Dick
aris at riponce dot com
27-Mar-2001 06:27
27-Mar-2001 06:27
If you test a symbolic (soft) link with is_file() it will return true. Either use filetype() which always returns the correct type OR make sure that you FIRST test with is_link() before you do with is_file() to get the correct type.