php新人求指教 postj接收数据问题

先说下 本人 是个菜鸟  代码可能有问题 见笑 了前端代码：
  apiready = function(){        
                api.ajax({
                    url: 'http://192.168.1.127/ThinkPHP/index.php/Home/Index/index.html',
                    method: 'post',
                    data: {
                        values: { 
                            name: 'zhaofeng'
                        },
                        
                    }
                },function(ret, err){
                    if (ret) {
                         api.alert({msg:JSON.stringify(ret)});
                    } else {
                         api.alert({msg:JSON.stringify(err)});
                    }
                });后端代码
<?php
namespace Home\Controller;
use Think\Controller;
class IndexController extends Controller {
    public function index(){
            $name = $_POST;
            $name_j = json_decode($name);
                        echo "$name";        //  也不知道怎么输出它 就都试试
                        dump($name);       //  也不知道怎么输出它 就都试试
                //        $this->display();
           if（$ret）{
                 $ret['flag']=100;$this->ajaxReturn($ret);
          }else{
                 $ret['flag']=101;$this->ajaxReturn($ret);}     总是返回这个101
   获取不到name啊 到底该怎么写呢
客户端返回的总是101                //判断用户是否登录过 通过session
                //if(isset($_SESSION['username']) && $_SESSION['username']!=''){                //}else{
                //        $this->redirect('User/index');
                //}
        }
}
public function index(){
            $name = $_POST['values'];
            $name_json = json_decode($name);
            $string = var_dump($name_json);
            echo "$string";
            if($ret){
                        $ret['flag']=100;$this->ajaxReturn($ret);
                }else{
                        $ret['flag']=101;$this->ajaxReturn($ret);
                }页面打输出 ：
<pre class='xdebug-var-dump' dir='ltr'><font color='#3465a4'>null</font>
</pre>{"flag":101}
  到底是怎么回事啊 为什么获取不到post传来的 name呢
下面是图片：