PHP: How to list files of a folder in a dropdown and then select one of them according to mysql entry

Question

I'm trying to create a dropdown listing the content of a specific folder. For that I'm using this code which works fine:

<select name="level">
<?php
    $dirname = "../images/page_images/";
    $dirhandle = opendir($dirname);
    while($file = readdir($dirhandle))
    {
    if ($file != "." && $file != "..")
    {
    if (is_file($dirname.$file))
    {
    echo "<option value='" . $file .">" . $file . "</option>"; 

    }
    else
    {
    echo "mappe: " . $file . "<br>";
    }
    }
    }
     ?> 
</select>

Now I want the dropdown bar to check a mysql entry and match it so that the file written in the database is the one that is selected. For that I think my option value should luke something like this: ($pageImage is the value loaded from mysql)

  <option <?php echo ($pageImage) == $file ? "selected" : "" ?> value="$file">$file</option>

My question is, how do I merge these two scripts together?

Answer 1

Try this one easy way to list file name from a folder.

<select name="level">
    <?php
    foreach (glob("../images/page_images/*.{jpg,gif}") as $filename) {
        echo "<option value='" . $filename .">" . $filename . "</option>"; 
    }
    ?>
    </select>

Answer 2

echo "<option value='{$file}'" . ($pageImage == $file ? " selected" : "") . ">{$file}</option>";

or

echo '<option value="' . $file . '"' . ($pageImage == $file ? " selected" : "") . '>' . $file . '</option>';

Docs: PHP String Operators