PHP: How to copy all images in a subfolder into a single folder (no need to create subfolders!)

Question

For my wife's birthday party, I set up a website through which I collected images of party guests and then created a nice book as a keepsake. Guests have an account and upload images to "their" folder. I now have a tool that dynamically creates slideshows from pictures - but unfortunately it cannot traverse subfolders. So my main goal is to copy the image file to a specified folder from which the slideshow can take it. I would then run a PHP script as cron every 5 minutes or so and display the image on the screen for the duration of the party.

I've found a bunch of code snippets that all do the same thing: They copy all files and folders recursively to the defined destination. For example. This one (taken from here: https://code-boxx.com/copy-folder-php/):

<?php
// (A) COPY ENTIRE FOLDER
function copyfolder ($from, $to, $ext="*") {
  // (A1) SOURCE FOLDER CHECK
  if (!is_dir($from)) { exit("$from does not exist"); }
 
  // (A2) CREATE DESTINATION FOLDER
  if (!is_dir($to)) {
    if (!mkdir($to)) { exit("Failed to create $to"); };
    echo "$to created\r\n";
  }
 
  // (A3) GET ALL FILES + FOLDERS IN SOURCE
  $all = glob("$from$ext", GLOB_MARK);
  print_r($all);
 
  // (A4) COPY FILES + RECURSIVE INTERNAL FOLDERS
  if (count($all)>0) { foreach ($all as $a) {
    $ff = basename($a); // CURRENT FILE/FOLDER
    if (is_dir($a)) {
      copyfolder("$from$ff/", "$to$ff/");
    } else {
      if (!copy($a, "$to$ff")) { exit("Error copying $a to $to$ff"); }
      echo "$a copied to $to$ff\r\n";
    }
  }}
}
 
// (B) GO!
copyfolder("C:/SOURCE/", "C:/TARGET/");
?>

This works fine, but the truth is it's not what I need. The script is copying files and folders and placing the files into the same subfolder they are in. My problem is that I don't want to create subfolders. I just want the script to go through all subfolders and copy the found image files into a folder. I thought this should be an easy thing for newbies, but it seems I was wrong.

Can anyone help me achieve this? Thanks!

Answer 1

IT Goldman was right - leave $to out of the equation; I even removed the $to variable entirely and put my path in the copy parameter:

0) { foreach ($all as $a) {
    $ff = basename($a); // CURRENT FILE/FOLDER
    if (is_dir($a)) {
      copyfolder("$from$ff/");
    } else {
      if (!copy($a, "C:/TARGET/$ff")) { exit("Error copying $a to C:/TARGET/$ff"); }
      echo "$a copied to C:/TARGET/$ff\r\n";
    }
  }}
}
 
// (B) GO!
copyfolder("C:/SOURCE/");
?>