当它被命名为'original.jpg'时，PHP无法在DIR中找到第一个文件

Question

I have a simple function that returns a valid image path to display. It is passed the URL that's stored for a particular row in my DB. The basic functionality is:

• If the URL has a trailing slash it's a directory; return the first file in that directory. If there aren't any, return the "default image"
• If the URL is an image, see that it exists, otherwise use the first rule.

It works perfectly, except if the folder only contains a file named 'original.jpg', it displays the default image. If I add another file, it can use 'original.jpg'. If I rename it to 'original.jpeg' or 'ori.jpg' (or shorter) it works. This is the only filename I've encountered that behaves this way.

function displayFile($file){
    $imgPath = "./img/path/";

    // If folder was specified or file doesn't exists; use first available file
    if( substr($file, -1) == '/' || !file_exists($imgPath . $file) ){
        // Extract base path
        $file = strstr( $file, '/', true );
        $handle = opendir($imgPath . $file);
        $entry = readdir($handle);
        $firstFile = '';
        while (false !== ($entry = readdir($handle))) {
            if( substr($entry, 0, 1) != '.' ){
                $firstFile = $entry;
                break; // This break isn't the problem
            }
        }
        // No file found; use default
        if( $firstFile == '' ){ return $imgPath . "paw.png"; }
        // Found a file to use
        else{ return $imgPath . $file . '/' . $firstFile; }
    } else {
        // File name is valid; use it
        return $imgPath . $file;
    }
    closedir($imgPath);
}

Answer 1

You're calling readdir twice, essentially always skipping the first entry.

$entry = readdir($handle);

Remove that line and you should be good to go.