I have a script where i upload files and then they are printed in a dropdown menu for me to add them to db; along with other information.

The code to print dropdown menu is as follows:

PHP Code:

<?php
                
                $dir = opendir('../images/upload');
                    while(false !== ($file = readdir($dir))){
                        if($file != "." && $file != ".." && $file != "Thumbs.db"){
                            echo '<option value="'.$file.'">'.$file.'</option>';
                        }
                    }
                    closedir($dir);
                    
                    
                ?>

Any chance of an if statement that will check if $file is in field (location), in table (games), in any record before printing

PHP Code:

<option value="'.$file.'">'.$file.'</option> 


?

Also if possoble, if it passes this test, it produces an array and sorts it, then prints it?

PHP Code:

<?php
$files = Array();
$dir = opendir('images');
while(false !== ($file = readdir($dir))){
  if($file != "." && $file != ".." && $file != "Thumbs.db"){
    $files[] = $file;
  }
}
closedir($dir);

$files=array_flip($files);

//Connect to your database here unless you have already

$result = mysql_query("SELECT location FROM games");

while($row = mysql_fetch_array($result)) {
  unset($files[$row['location']]);
}

$files= array_flip($files);

sort($file);
foreach($files as $file) {
  echo '<option value="'.$file.'">'.$file.'</option>';
}


?>

Try this ^

You sorted $file rather than $files, but i sorted that and costomised it and it works now.

THANKS

Ooops, well spotted.