Count Files in path /

duny

Is there a line of code out there that will count the number of files in my path /

bubblenut

Recursively or just in / ?

duny

Recursively.....through out all of my dirunder the /

bubblenut

function file_lister($path) {
  $output=array();
  if(is_dir($path)) {
    $dir=opendir($path);
    while($file=readdir($dir)) {
      if(is_dir($file)) {
        array_push($output,file_lister($file));
      } else {
        $output[]=$file;
      }
    }
  }
  return $output;
}

$output=file_list('/');

I haven't tested it at all and I can think of al least two things which may be wrong but hte logic is about right.
I wouldn't say this is a very good thing to do at all thought as if you have any large number of files this will take f****g ages to run, and probably run out of memory pretty quickly. Think, if you've got say a 40GB HDD and on nix machines there are a lot of tiny files all over the place.
I would say is create some abstaction layer and then use native commands (ls -R on *nix, could even do ls -R > file and then just read the file line by line) using shell_exec.
HTH
Bubble

duny

Thanks is there one that would be faster just to say just the amount of files...

i.e. (there are 123 files in this web site.)

bubblenut

function file_lister($path) {
  $output=0;
  if(is_dir($path)) {
    $dir=opendir($path);
    while($file=readdir($dir)) {
      if(is_dir($file)) {
        $output+=file_lister($file);
      } else {
        $output++;
      }
    }
  }
  return $output;
}

$output=file_list('/');

When you say / do you mean the root of the whole system or the root of your website. You shouldn't be storing your website in / (although it may be set up to appear as such if you're working on someone elses server)
Bubble

Weedpacket

Originally posted by duny
Thanks is there one that would be faster just to say just the amount of files...

i.e. (there are 123 files in this web site.)

Yes. Count the files found instead of putting them in an array.

Or, as bubblenut notes, given typical sizes, do it with a system call (ls -U1Rd and pipe the result through something to strip out blank lines and lines ending with =:, then count the number of lines left). Um ls -U1Rd | egrep -cv "=:$|^$ I think. You won't be able to go much faster than asking the system itself.

(Er, but do you want the entire hard drive, or just a web document tree?)