Untitled

<html>
<head></head>
</body>
<form id="myform" action="formdata.php" method="post">
    username:<input type="text" name="username" id="name"><br>
    password:<input type="password" name="password" id="pass"><br>
    firstname:<input type="text" name="firstname" id="fname"><br>
    lastname:<input type="text" name="lastname" id="lname"><br>
    <input type="submit" id="submit" value="register">
</form>
<div id="status_text"></div>
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.7.1/jquery.min.js"></script>
<script>
$("#submit").click(function(){
    var username = $('#name').val();
    var password = $('#pass').val();
    var firstname = $('#fname').val();
    var lastnamee = $('#lname').val();
    var postData = '&username='+username+'&password='+password+'&firstname='+firstname+'&lastname='+lastname;
    var status_text = $('#status_text');
    //alert(postData);
    //var mydata = {'username':name,'password':pass,'firstname':fname,'lastname':lname};
/*$.post($('#myform').attr('action'),
    $('#myform:input').serializeArray(),
    function(info){
        $('status_text').html(info)
    });*/
$.ajax({
    url:"action",
    type:"post",
    success:function(info)
    {
        status_text.html(info);
    }
});
});
</script>
</body>
</html>

<?php
$servername = 'localhost';
$username = 'root';
$password = '';
$conn = new mysqli($servername, $username, $password);
$name = $_POST['username'];
$password = $_POST['password'];
$firstname = $_POST['firstname'];
$lastname = $_POST['lastname'];
$sql = "INSERT INTO karthik.person(name,password,firstname,lastname)VALUES('$name','$password','$firstname','$lastname')";
if($conn->query($sql) === TRUE){
    echo("record added success");
}else{
    echo("failed".$conn->error);
}
$conn->close();
?>