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 | I know there are many other threads about this same error, but none have helpped me, here is my code: [CODE] <?php $mysql_id = mysql_connect('mysql3.000webhost.com', 'a2778852_556875', 'pendolino390'); mysql_select_db('a2778852_555676', $mysql_id); $result = mysql_query("SELECT personalexperience, sex, age, sexuality, FROM Personal_experience"); while($row = mysql_fetch_array($result)) { echo $row['personalexperience'], $row['sex'], $row['age'], $row['sexuality']; } mysql_close($mysql_id); ?>[/CODE] …  |
| | Hi I'm Carrick, Basically, I'm a complete novice when it comes to web design and I decided to join this site to see whether it could help me get better as it were.  |
| | Whilst working on my first MYSQL, i came accross this error, and even with a friend who is a pretty good coder and we couldn't work it out [CODE]<?php $con=mysql_connect("mysql3.000webhost.com","a2778852_556875", "pendolino390"); //CREATE DATABASE mysql_create_db("personal_experience", $con); //CREATE TABLE $con=mysql_select_db('Personal_experiences',$con); if($con) { $sql="CREATE DATABASE personal_ex ( Additioncode int NOT NULL AUTO_INCREMENT, PRIMARY …  |