I'd pull the addslashes() and use mysql_real_escape_string() instead. I'd also echo the query that's being sent to the database and compare to make sure it's doing what it should. -Micah On Thursday 10 November 2005 8:22 pm, JeRRy wrote: > Hi, > > Well I tried this code but it fails, if I enter a correct User and Pass set > in the db it just prompts for the user/pass again. The error message that > should apply if wrong three times appears but the sucessful message I can't > get regardless of correct user pass or not, any help please? > > <?php > function displayLogin() { > header("WWW-Authenticate: Basic realm=\"My Website\""); > header("HTTP/1.0 401 Unauthorized"); > echo "<h2>Authentication Failure</h2>"; > echo "The username and password provided did not work. Please reload this > page and try again."; exit; > } > $db = mysql_connect('localhost','db_user',db_pass') or die("Couldn't > connect to the database."); mysql_select_db('db_name') or die("Couldn't > select the database"); if (!isset($PHP_AUTH_USER) || !isset($PHP_AUTH_PW)) > { > // If username or password hasn't been set, display the login request. > displayLogin(); > } else { > // Escape both the password and username string to prevent users from > inserting bogus data. $PHP_AUTH_USER = addslashes($PHP_AUTH_USER); > $PHP_AUTH_PW = md5($PHP_AUTH_PW); > // Check username and password agains the database. > $result = mysql_query("SELECT count(id) FROM users WHERE > password='$PHP_AUTH_PW' AND username='$PHP_AUTH_USER'") or die("Couldn't > query the user-database."); $num = mysql_result($result, 0); > if (!$num) { > // If there were no matching users, show the login > displayLogin(); > } > } > // All code/html below will only be displayed to authenticated users. > echo "Congratulations! You're now authenticated."; > > ?> > > > J > > > --------------------------------- > Do you Yahoo!? > Yahoo! News: Get the latest news via video today! -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php