Regulary Expression for filtering redundant directory notations

Try so (perl):
$cur_dir="/../../../dir";
$cur_dir=~s/\/..//i;

Now, variable $cur_dir contain "/dir". In this part of a code all "/.." are replaced on "".
Thus I too am defended from hackers.

I'm sorry.
Valid code is:
$cur_dir="/../../../dir";
$cur_dir=~s/\/..//g;     # "g" instead of "i"

Damn!!! I only need the expression, not the Perl code! ;-)

Sorry, but your solution does not work in php PREG_xxx-command. ~s seems to be PERL but not part of a PERL regulary expression?!

222 if you give me a single common Perl RegExpr that works with php preg_replace( )-function (PHP uses the PERL module)!

s#\.\./##g

Your regulary expression only reduces all repeating /.. to one /.. It would shorten

   /part1/part2/../..

to

   /part1/part2/..

But I don't want to cut the end of the string but kill the redundant directories notations from the path. I want the "true" path in this example it has to be:

/    (base directory)

as /part1/part2 and /../.. are compensating each other!

hope now it is now more clearly what I want! Is there still someone who can help me!? :-)

what you want is not a regex but a regex-substitution:
 what do you expect for: /part1/part2/../../xx/../../..

#Get rid of "/./"  to handle /./..
$cur_dir =~ s{/\./}{/}g;

#Get rid of "//"
$cur_dir =~ s{//}{/}g;

#Simplify /dir/../ patterns to /
my ($temp) = "";
while ($temp ne $cur_dir) {
    $temp = $cur_dir;
    $cur_dir =~ s{/[^./][^/]+/\.\.(/|$)}{/};  # Handle most cases
    $cur_dir =~ s{/\.[^.][^/]+/\.\.(/|$)}{/}; # Handle most cases with directories starting with "."
    $cur_dir =~ s{/\.\.[^/]+/\/.\.(/|$)}{/};   # Handle the last few cases with dirs starting with ".."
}

# Clean up a trailing / that may have been added
$cur_dir =~ s{/$}{};

Since you need to backtrace after each replacement, you can't use a "g" modifyer, but instead you must use a loop.

@ahoffmann:

expecting function that generates from

/part1/part2/../../xx/../../..

the result path is

"/"

or ε (empty string!)

(deepest allowed path is the base directory "/" - not nescessarily the root directory of the filesystem but the deepest allowed directory. In my case it is the directory from which the script is called.)

>> deepest allowed path is the base directory "/" - not nescessarily the root directory

I don't know how php "reads" path notation, but on unix systems / is the root dir and ./ is the current working directory (or "base directory").

So far, each of the solutions removes the ../ relative path notation but may leave you with a properly formatted but a non existing directory path.   Also nothing has been mentioned about the user inputting an absolute path that is higher up the dir tree than the previous path.  Following along with the relative path problem (as outlined in your last post), you could do this (using Perl syntax since I don't know php):

$cur_dir = './' if ($cur_dir =~ /\.\./);

or this:

$cur_dir =~ s#^.*?\.\./.*$#./#;

If you need to test for absolute paths, then we'll need to approach this from a slightly different angle.

In Perl-speak, I'd use something like this:

my $path = '/part1/part2/../../xx/../../../';
1 while($path =~ s#(?:[^/]+/)?\.\./##);

And it php, it'd look more like this:

$path = '/part1/part2/../../xx/../../';
while($path != ($tmp = preg_replace( "#(?:[^/]+/)?\.\./#", "", $path, 1)))
        $path = $tmp;

And people say Perl isn't beautiful. :) If you have a trailing / on the path, it leaves you with that, otherwise, it ends with the empty string.

icrf, I came up with a similar regex:
  while (s#[^/.]+[/]+\.\./##){}

but it suffers from the same problem as your sugestion, use
   $path = '/part1/part2/../../xx/../../..';

Think I need to go to bed with J.Friedl's regular ex. and a few beers ...
Probably there is something better with perl's look-ahead tomorrow ..

Okay...
ahoffmann's preg-expression works perfectly! :-)
Thank you to icrf too!

working php solution for eliminating redundant directories in pathes is:

$path = '/part1/part2/../../xx/../../..';
while($path != ($tmp = preg_replace( "#(?:[^/]+[/]+)+\.\./?#", "", $path, 1))) $path = $tmp;
echo $path;