Question

I’ve been making a form recently. It works fine. I just want to add one tweak to it that it shows an error if one registers an email id that has already been registered in the database.
Here’s my whole php code::

<?php
      //include the connection file

      require_once("validation.php");

if( isset($_POST['send']) && (!validateName($_POST['name']) || !validateEmail($_POST['email']) || !validatePasswords($_POST['pass1'], $_POST['pass2']) || !validateContact($_POST['contact']) || !validateAge($_POST['age'])) ):?>
            <div id="error">    
                <ul>
                    <? if(!validateName($_POST['name'])):?>
                        <li><strong>Invalid Name:</strong> We want names with more than 3 letters.</li>
                    <? endif?>
                    <? if(!validateEmail($_POST['email'])):?>
                        <li><strong>Invalid E-mail:</strong> Type a valid e-mail please.</li>
                    <? endif?>
                    <? if(!validatePasswords($_POST['pass1'], $_POST['pass2'])):?>
                        <li><strong>Passwords are invalid:</strong> Passwords doesnt match or are invalid!</li>
                    <? endif?>
                    <? if(!validateContact($_POST['contact'])):?>
                        <li><strong>Please enter a valid number.</strong></li>
                    <? endif?>
                    <? if(!validateAge($_POST['age'])):?>
                        <li><strong>Please enter a valid age</strong></li>
                    <? endif?>
                    <? if(!validateWork($_POST['work'])):?>
                        <li><strong>Please enter work</strong></li>
                    <? endif?>
                    </ul>
            </div>
        <?php elseif(isset($_POST['send'])):?>
            <div id="error" class="valid">
                <ul>
                <?php
                require_once('connection.php'); 
                $query = "INSERT INTO employee (name, password, email, contact, age, gender, location, skill, work) VALUES ";                           
                $query .= "('".$_POST['name']."', '".$_POST['pass1']."', '".$_POST['email']."','".$_POST['contact']."','".$_POST['age']."','".$_POST['gender']."','".$_POST['location']."','".$_POST['skill']."','".$_POST['work']."')";
                // run the query
                mysql_query($query);?>
                <li><strong>Congratulations!</strong> You have been successfully registered!</li>
                </ul>
            </div>
    <?php endif?>

Answer 1

first, make your email a unique one:

CREATE TABLE employee { ..., UNIQUE KEY theemail (email) }   # in create table

or via query:

ALTER TABLE employee ADD UNIQUE KEY theemail (email);   # via console etc. phpmyadmin

then, you can just check the mysql_affected_rows as @sachleen said above:

$query = "INSERT INTO employee (name, password, email, contact, age, gender, location, skill, work) VALUES ";                           
$query .= "('".$_POST['name']."', '".$_POST['pass1']."', '".$_POST['email']."','".$_POST['contact']."','".$_POST['age']."','".$_POST['gender']."','".$_POST['location']."','".$_POST['skill']."','".$_POST['work']."')";
$result = mysql_query($query);
$affected_rows = ($result ? mysql_affected_rows($result) : 0);
if ($affected_rows==0) {
   // fail
}

ot more simply:

$result = mysql_query($query);
if (!$result) {
   // fail
}