我正在使用Cake PHP v2.2.1稳定版.我有一个带有动作add()的UsersController.我正在尝试通过ajax(从主页到/ users / add)发送用户信息并保存数据.我的代码是这样的：

// /app/View/Pages/home.ctp
<?PHP
   $data = array('User' => array('username' => 'vegeta_suPEr_sayajin','password' => 'over9000!','email' => 'vegeta@supersayajin.COM','PRofile_pic' => '/home/pics/scouter.jpg','Firstname' => 'Vegeta','lastname' => 'Vegeta','level_id' => '9001'));
?>
<script type="text/javascript">
    VAR data = <?PHP echo json_encode($data); ?> //convert $data into JSON format
    $.ajax({url: '/users/add',data: "data="+data,type: 'post'});
</script>

如何在UsersController中接收这些数据,以便我可以处理和保存它？

目前,我正在尝试：

// /app/Controller/UsersController.PHP
function add() {
    if($this->request->is('post') {
        //returns "Error: [object Object] in LOGfile
        $this->log($this->request->data);
    } else {
        $this->Session->setFlash(__("The user Could not be saved :("));
    }
    $this->autoRender = false;
}

$this-> log($this-> request-> data)在/app/tmp/logs/error.log文件中返回错误：[object Object],并且$user中的任何一个都不存在此用户信息this-> request-> params的索引.到目前为止,我所有的谷歌搜索都只返回了复杂的cakePHP v1.3技术.在cakePHP v2.2.1中如何完成？

<?PHP
$data = array(
'User' => array(
    'username' => 'vegeta_super_sayajin','level_id' => '9001')
    );   
?>
<script type="text/javascript">
var data = [<?PHP echo json_encode($data); ?>] //convert $data into json format
$.ajax({
    url: 'checks/add',data: "data="+JSON.stringify(data),type: 'post'});
</script>

// /app/Controller/UsersController.PHP
function add() {
if($this->request->is('post') {
    $this->log(json_encode($this->request->data,true)); //returns "Error: [object Object] in logfile
} else {
    $this->Session->setFlash(__("The user Could not be saved :("));
}
$this->autoRender = false;
}