ajax - 通过ajax发送数据时Codeigniter 403 Forbidden Error

Question

我已经使用 Codeigniter 3 构建了一个简单的项目，当我在本地主机中使用 $.ajax 方法发送 ajax 请求时，它运行良好，但我得到了

"403 Forbidden"

当我在实时服务器上这样做时出错。我在 config.php 中设置了 $config['csrf_protection'] = FALSE; 和 $config['csrf_regenerate'] = FALSE;。这是使用ajax发送数据的js代码。

     $.ajax({
        url : '/login/authenticate',
        type : 'post',
        data : $(this).serialize(),
        success : function(response) {
            if (response.state == false) {
                var msg = response.msg;
                err_msg(msg);
            } else {
                if(response.type == "admin"){
                    window.location.href ='/admin';    
                } else {
                    window.location.href = '/user';
                }
            }
        }
    });

请告诉我如何解决这个问题。这是我的登录 Controller

class Login extends CI_Controller {

public function __construct()
{
    parent::__construct();
}

public function index()
{
    $this->load->view('login_view');
}

//check the email and password and log the user in if the user info is correct
public function authenticate()
{   
    $this->load->model("userModel","user", true);
    $this->load->library('form_validation');
    //Form validation - codeigniter provides you with powerful form validation functionality
    $email = $this->input->post('email');
    $password = $this->input->post('password');

    $this->form_validation->set_rules('email', 'Email', 'trim|required|valid_email');
    $this->form_validation->set_rules('password', 'Password', 'required');

    if ($this->form_validation->run() == FALSE) {
        $res = array('state' => false, 'msg' => validation_errors());
    } else {
        $type = $this->user->login($email, $password);
        if ($type == "user" ) {
            $res   = array('state' => true, 'type' => $type, 'msg' => 'You are logged in!');
            $toast = array('state' => true, 'msg' => 'You are logged in!');
            $this->session->set_flashdata('toast', $toast);

        }else if($type == "admin"){
            $res   = array('state' => true, 'type' => $type, 'msg' => 'You are logged in!');
            $toast = array('state' => true, 'msg' => 'You are logged in!');
            $this->session->set_flashdata('toast', $toast);
        }else if ($type == -3) {
            $msg = "You can't be logged in because you are not active at the moment.";
            $res = array('state' => false, 'msg' => $msg);
        }else if ($type == -1) {
            $msg = "Wrong Password!";
            $res = array('state' => false, 'msg' => $msg);
        }else {
            $msg = "You were not registered!";
            $res = array('state' => false, 'msg' => $msg);
        }
    }
    return $this->output
                ->set_content_type('application/json')
                ->set_output(json_encode($res));        
}
}

Answer 1

你好，我在使用上面的代码时没有遇到任何错误......见下图

使用表单名称或表单 ID 来序列化数据而不是 $(this).serialize()