thinkphp3.2 ajax 提交返回数据

html页面代码：

<div>
    <form method="post">
        <input type="text" name="username" />
        <input type="password" name="password" />
        <button type="submit">Submit</button>
    </form>
    <p id="content"></p>
</div>


<script src="http://libs.baidu.com/jquery/2.0.0/jquery.min.js"></script>
<script>
    $(function() {
        $("button[type=submit]").on("click", function() {
            $.ajax({
                type: "POST",
                url:"{:U("home/timu/checkLogin")}",
                data: {
                    username: $("input[name=username]").val(),
                    password: $("input[name=password]").val()
                },
                success: function($data) {
                    $("#content").html($data.info);
                }
            });
            return false;
        });
    });
</script>

控制器方法代码：

#ajax 请求！
    public function checkLogin(){

        if (IS_POST) {
            $data = I("post.");
            empty($data["username"]) && $this->error("用户名不能为空");
            empty($data["password"]) && $this->error("密码不能为空");
            $this->success("初步验证成功！");
        }
    $this->display();
    }