Re: Re: creating a text file

I have tried that, but I can't figure out how to then change my variable: $data_file = "pictures/pictures/list.photos"; to access the info in the script below or similar scripts. How can I get the variable $data_file to read the file names from the script below?

Try something like this: if ($dir = @opendir('/path/to/pics') {

    while (($file = readdir($dir)) !== false) {
        if(strpos($file, ".jpg")>0) {
	    $names[] = $file;
    	}
    }
    closedir($dir);
    sort($names); // optional

} HTH, Karsten -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php

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