I am unable to retrieve the value of $referral_1 from: $new_email = mysql_escape_string ( $_POST['referral_$i'] ); why? PHP while lopp to check if any of the fields were populated: $i=1; while ( $i <= 5 ) { $new_email = mysql_escape_string ( $_POST['referral_$i'] ); if ( strlen ( $new_email ) > 0 ) { .... } } The form itself: <form method="post" action=âwebsiteâ> <p style="margin: 10px 50px 10px 50px;"><input type="text" name="email" maxlength="60" class="referral_1" style="width: 400px;"></p> <p style="margin: 10px 50px 10px 50px;"><input type="text" name="email" maxlength="60" class="referral_2" style="width: 400px;"></p> <p style="margin: 10px 50px 10px 50px;"><input type="text" name="email" maxlength="60" class="referral_3" style="width: 400px;"></p> <p style="margin: 10px 50px 10px 50px;"><input type="text" name="email" maxlength="60" class="referral_4" style="width: 400px;"></p> <p style="margin: 10px 50px 10px 50px;"><input type="text" name="email" maxlength="60" class="referral_5" style="width: 400px;"></p> <p style="margin: 10px 50px 10px 50px;"><input type="submit" name="submit" value="Add New Referrals"></p> </form> What am I doing wrong? Ron The Verse of the Day âEncouragement from Godâs Wordâ http://www.TheVerseOfTheDay.info