CODEWITHSUNDEEP
phpjavascriptjquerymysqlajax
Some user
Some user

Reputation: 49

php check if a username exists in a database

So I have a user form registration, and what I am trying to do is: while a user is typing an email, the website will check my database if the email already has been used or not, before they hit the register button.

The problem I'm having is that it won't check. It will only display "Searching in database". I just want to post my code so maybe someone can catch the error I'm making.

This is part of my registration page:

<tr class = "spacearound"> <!-- input for email address -->
    <th> &emsp;Email: </th>
    <td>
        <input type = "text" id = "user_email" size = "50"
            maxlength = "50" name = "u_email" 
            title = "Enter your email please" 
            onchange = "EmailCheck();" 
            onkeypress = "return InputLimiter(event, 'emailCharacters');"
            /> &#42;
        <span id = "email_status"> </span>
    </td>
    <td><?php  echo $message; ?></td>
</tr>

This is my JavaScript file, "checkusers.js":

$('#user_email').keyup(function() {
    var username = $(this).val();

    $('#email_status').text('Searching database.');

    if(username != ''){
        $.post('checkemail.php',{ username: username }, function(data) {
            $('#email_status').text(data);
        });
    } else {
        $('#email_status').text('');
    }

});

And this is my php file, where I check for an email, "checkemail.php":

<?php
    define('dbHost', 'xxxxx');
    define('dbUser', 'xxxxx');
    define('dbPassword', 'xxxxx');
    define('dbName', 'xxxxx');
    error_reporting(E_ALL ^ E_NOTICE);

    $db = mysqli_connect(dbHost, dbUser, dbPassword, dbName);

    if(mysqli_connect_errno()) { //if connection database fails
        echo("Connection not established ");
    }  //by now we have connection to the database

    if(isset($_POST))['username'])){ //if we get the name succesfully
    $username = mysqli_real_escape_string($db, $_POST['username']);
        if (!empty($username)) {
            $username_query = mysqli_query($db, "SELECT COUNT(`firstName`) FROM `users` WHERE `email`='$username'");    

        $username_result = mysqli_fetch_row($username_query);

            if ($username_result[0] == '0') {
                echo 'Email available!';
            } else {
                echo 'Sorry, the email '.$username.' is taken.';
            }
        }
    }

?>

Upvotes: 1

Views: 14655

Answers (1)

kashimu
kashimu

Reputation: 174

you have error here 

if(isset($_POST))['username'])){

it should be

if(isset($_POST['username'])){

$_POST['username'] should be enclosed inside isset function

Upvotes: 4

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