How can i make the below code work!It's a connection to the db.Whene a user tries to sign up i want to display an error if their email input exists try another one but if its not then you are signed up!
$query = "SELECT * FROM `TesteExample` WHERE Email_address = '".mysqli_real_escape_string($link,$_POST['email_address'])."'";
$resultQuery = mysqli_query($link,$query);
$resultsQuery = mysqli_num_rows($resultQuery);
//We test if the email the user tries is already registered.If it is then we tell him to create one.
if($resultsQuery) {
$AlreadyExists = "There is already a user with this email address.Please try another one.";
} else {
$query = "INSERT INTO `TesteExample`
(`Name`, `Surname`, `Email_address`, `Password`, `Gender`, `Comment`, `Country`, `ListOfDates`, `ListOfMonths`, `ListOfYears`, `Website`)
VALUES ('".$_POST['name']."','".$_POST['surname']."','".mysqli_real_escape_string($link,$_POST['email_address'])."','".$_POST['password']."','".$_POST['gender']."','".$_POST['comment']."','".$_POST['country']."','".$_POST['listOfDates']."','".$_POST['listOfMonths']."','".$_POST['listOfYears']."','".$_POST['website']."')";
mysqli_query = ($link,$query);
Notitce that is i comment out the else statement everything works fine.But if i include it the whole page doesnt work.. I want you to know that some of them like(country) is data from a dropdown list or (gender) is a radio button. Do i have to specify anything else at the INSERT INTO ????