Filesystem 函数
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PHP手册
Filesystem 函数
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PHP手册
PHP手册 - N: 判断给定文件名是否为一个符号连接

用户评论:

PaulE (02-Jun-2009 10:03)

A workaround for Windows LNK files:

<?php

function _is_link($filename)
{
    if(is_link($filename))
        return true;

    $ext = substr(strrchr($filename, '.'), 1);
    if(strtolower($ext) == 'lnk')
    {
        return (_readlink($filename) ? true : false);
    }

    return false;
}

function _readlink($file)
{
    if(file_exists($file))
    {
        if(is_link($file))
        {
            return readlink($file);
        }

        // Get file content
        $handle = fopen($file, "rb");
        $buffer = array();

        while(!feof($handle))
        {
            $buffer[] = fread($handle, 1);
        }

        fclose($handle);

        // Test magic value and GUID
        if(count($buffer) < 20)
            return false;
        if($buffer[0] != 'L')
            return false;
        if((ord($buffer[4]) != 0x01) ||
           (ord($buffer[5]) != 0x14) ||
           (ord($buffer[6]) != 0x02) ||
           (ord($buffer[7]) != 0x00) ||
           (ord($buffer[8]) != 0x00) ||
           (ord($buffer[9]) != 0x00) ||
           (ord($buffer[10]) != 0x00) ||
           (ord($buffer[11]) != 0x00) ||
           (ord($buffer[12]) != 0xC0) ||
           (ord($buffer[13]) != 0x00) ||
           (ord($buffer[14]) != 0x00) ||
           (ord($buffer[15]) != 0x00) ||
           (ord($buffer[16]) != 0x00) ||
           (ord($buffer[17]) != 0x00) ||
           (ord($buffer[18]) != 0x00) ||
           (ord($buffer[19]) != 0x46))
        {
            return false;
        }

        $i = 20;
        if(count($buffer) < ($i + 4))
            return false;

        $flags = ord($buffer[$i]);
        $flags = $flags | (ord($buffer[++$i]) << 8);
        $flags = $flags | (ord($buffer[++$i]) << 16);
        $flags = $flags | (ord($buffer[++$i]) << 24);

        $hasShellItemIdList = ($flags & 0x00000001) ? true : false;
        $pointsToFileOrDir = ($flags & 0x00000002) ? true : false;

        if(!$pointsToFileOrDir)
            return false;

        if($hasShellItemIdList)
        {
            $i = 76;
            if(count($buffer) < ($i + 2))
                return false;

            $a = ord($buffer[$i]);
            $a = $a | (ord($buffer[++$i]) << 8);
           
        }

        $i = 78 + 4 + $a;
        if(count($buffer) < ($i + 4))
            return false;

        $b = ord($buffer[$i]);
        $b = $b | (ord($buffer[++$i]) << 8);
        $b = $b | (ord($buffer[++$i]) << 16);
        $b = $b | (ord($buffer[++$i]) << 24);

        $i = 78 + $a + $b;
        if(count($buffer) < ($i + 4))
            return false;

        $c = ord($buffer[$i]);
        $c = $c | (ord($buffer[++$i]) << 8);
        $c = $c | (ord($buffer[++$i]) << 16);
        $c = $c | (ord($buffer[++$i]) << 24);

        $i = 78 + $a + $b + $c;
        if(count($buffer) < ($i +1))
            return false;

        $linkedTarget = "";
        for(;$i < count($buffer); ++$i)
        {
            if(!ord($buffer[$i]))
                break;

            $linkedTarget .= $buffer[$i];
        }

        if(empty($linkedTarget))
            return false;

       
        return $linkedTarget;
    }

    return false;
}

if(_is_link('test.lnk'))
{
    echo _readlink('test.lnk');
}
?>

foobarfarter (18-Oct-2008 11:06)

@radon8472 at hotmail dot com
The windows .lnk-things are real files, the explorer only treats them like links. Try to open one with Win+R->"notepad X:\Path\Visiblefilename.lnk You will see much 0-Bytes, but the linked path is displayed right on the screen. It should be possible to do the same by php.

radon8472 at hotmail dot com (01-Aug-2008 05:38)

This function returns "false" for windows *.lnk files (tested under windows XP).

filetype also returns "file" an not "link" like expected.

I didn`t find a solution for this.

mbirth at webwriters dot de (20-May-2008 04:24)

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.

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 03:22)

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 07:31)

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 06:44)

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.