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)
is_link
(PHP 4, PHP 5)
is_link — Zjistí, zda se jedná o symbolický odkaz (link)
Popis
bool is_link ( string $filename )Vrací TRUE, když soubor existuje a jedná se o symbolický odkaz (link).
Výsledek této funkce je cachován. Více detailů - viz clearstatcache().
Viz také is_dir() a is_file().
Poznámka: Tato funkce nefunguje na Windows systémech
is_link
neverpanic->gmail[com]
30-Dec-2006 02:50
30-Dec-2006 02:50
brendy at gmail dot com
05-May-2006 07:22
05-May-2006 07:22
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 ( $target == "" ? false : true );
}
?>
jr at cnb dot uam dot es
30-May-2005 11:31
30-May-2005 11:31
Why don't you just try
is_dir("$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
02-Jun-2002 10:44
02-Jun-2002 10:44
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 02:27
27-Mar-2001 02:27
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.
