Question

我正在开发一个Web应用程序。该应用程序正在使用Laravel Framework，并且应该通过另一个已经存在的REST-API获取所有数据，以及有关已认证用户的信息。该REST-API使用基本身份验证作为确认身份验证的方法。

我到目前为止所做的是以下代码：

function userLogin(){

    //get username and password from login form
    $username = $_POST["username"];
    $password = $_POST["password"];

    //create a new Guzzle client
    $client = new Client();

    //send username and password to REST API to get the Authentication
    $response = $client->get('localhost:8080/api/user/login', [
        'auth' => [$username, $password]
    ]);

    if ($response->getStatusCode() < 200 || $response->getStatusCode() >= 300) {
        return back()->with('danger','Login failed.');
    }else{
        return view('home');
    }
}

我设法登录，但是当我单击菜单时，由于中间件不知道用户已登录，它使我回到“登录”页面。

我还从同事那里获得了一个Angular代码，该代码已成功在他的Ionic Mobile Apps中成功管理了登录尝试：

login(user:User): Observable<any> {
const headers = new HttpHeaders(user ?
  {authorization:'Basic ' + btoa(user.username + ":" + user.password)}
  :
  {});

  /**
   * a simple temporary measure to counter the error we get
   * when we log in to the app. 
   * because the API sends nothing as response, nothing is assigned to the currentUser variable.
   * this one line only assign value of currentUser manually to localStorage.
   * it consists only of username and password.
   */
  localStorage.setItem('currentUser', JSON.stringify(user));

  /**
   * as of now these do nothing, because the API sends out nothing as response.
   */
  return this.http.get<any>(API_URL + "login", {headers: headers})
  .pipe(map(response => {
    if(response){
      localStorage.setItem('currentUser', JSON.stringify(response));
    }
    return response;
  }));
}

不幸的是，我不知道如何将其转换为PHP代码或Laravel函数。我希望有人可以给我解决这个问题的方法。

Answer 1

在后端进行身份验证时，您必须应用中间件来对用户进行身份验证，因此必须在Authenticate.php或任何其他自定义中间件中编写代码：

try{

        $client = new \GuzzleHttp\Client();
        $url = "https://identitytoolkit.googleapis.com/v1/accounts:lookup?key=".env("API_KEY");

        $request = $client->post($url, array(
            'form_params' => array(
                'idToken' => $token
            )
        ));
        //$response = $request->getBody();
        //$response = $request->send();
        $body = $request->getBody();

        if ($request->getStatusCode() !== 200) {
            $res['status'] = 401;
            $res['message'] = 'Error, Invalid domain.';
            return response($res);
        }
    }
    catch(\GuzzleHttp\Exception\ClientException $e) {
       log::info('catch========== Exception');
        $res['status'] = 401;
        $res['message'] = 'Error, Unauthorized.';
        return response($res);
    }

来自第三方API的Laravel基本身份验证

1 个答案: