使用$ .ajax与PHP进行JQuery

I've been struggling for the entire day, I can't seem to get it right to send through a JSON object to a PHP file using the $.ajax function. I eventually got it right to send it through but now it refuses to decode.

//JQuery File

            var user = 
        {
            "email" : document.getElementById('email1').value,
            "password" : document.getElementById('pwd1').value,
            "fname" : document.getElementById('firstname').value,
            "lname" : document.getElementById('lastname').value,
            "gender" : Validator.getGender(),
            "dob" : document.getElementById('dob').value
        }; 
       JSON.stringify(user);

            user = {json:user};

            $.ajax({
                    type: "POST",
                    url: "register.php",
                    dataType: 'json',
                    data: user,
                    success: function(result)
                    {       

                        alert("It worked :D");
                        alert(result);
                    },
                    failure: function()
                    {

                        alert('whoops');
                    }
                });

//PHP File

<?php
    $json = $_REQUEST['json'];
    $json = stripslashes($json);

    $jsonobj = json_decode($json);

    $fname = $jsonobj->fname;
    $lname = $jsonobj->lname;
    $password = $jsonobj->password;
    $email = $jsonobj->email;
    $gender = $jsonobj->gender;
    $dob = $jsonobj->dob;
    echo $gender;

?>

Am I doing something completely silly ? I'm just echoing the gender variable back for now but am actually going to use this to enter it into a database.

Thanks in advance :)

EDIT:

How would it would it work if my php file looked like this:

    session_start();
$db = mysql_connect("localhost", "root");
if(!$db)
{
    die("DB connection failed: " . mysql_error());
}

$db_select = mysql_select_db("tinyspace", $db);
if(!$db_select)
{
    die("DB connection failed: " . mysql_error());
}

$json_string = $_REQUEST["json"];
$jsonobj = json_decode($_REQUEST["json"]);
$fname = $jsonobj -> fname;
$lname = $jsonobj -> lname;
$pwd = $jsonobj -> password;
$email = $jsonobj -> email;
$gender = $jsonobj -> gender;
$dob = $jsonobj -> dob;

$sql("INSERT INTO tinyspace.users (email, password, firstname, lastname, gender, dob) VALUES ('$email','$password','$fname', '$lname','$gender','$dob')");

if (!mysql_query($sql,$db))
{
    die('Error: ' . mysql_error());
}

mysql_close($db);

The problem is that you are using datatype =json and trying to echo result.So it won't work it will accept only json data

you need to remove this

dataType: 'json',

or echo someting like this

$gender=array();
$gender['gender']=$jsonobj->gender
echo json_encode($gender);

You are also not creating the json

use this

var user = 
        {
            "email" : document.getElementById('email1').value,
            "password" : document.getElementById('pwd1').value,
            "fname" : document.getElementById('firstname').value,
            "lname" : document.getElementById('lastname').value,
            "gender" : Validator.getGender(),
            "dob" : document.getElementById('dob').value
        }; 
      var userjson= JSON.stringify(user);

            user = {json:userjson};

or

data:{'json' :userjson}