is_link

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。

注意: 本函数的结果会被缓存。更多信息参见 clearstatcache()。

提示: 自 PHP 5.0.0 起本函数也可被某些 URL wrapper 使用。参考附录 N 来看哪些 wrapper 支持 stat() 系列函数的功能。

参见 is_dir()，is_file() 和 readlink()。

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neverpanic->gmail[com]
30-Dec-2006 10:50


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 02:22


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 06:31


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 05:44


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 10:27


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.

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