Question

因此，我尝试通过登录请求返回User数据。我正在使用食肉，但我不知道如何将数据附加到响应中。

这是登录方法

public function login(Request $request)
    {
        $http = new \GuzzleHttp\Client;

        try {
            $response = $http->post(config('services.passport.login_endpoint'), [
                'form_params' => [
                    'grant_type' => 'password',
                    'client_id' => config('services.passport.client_id'),
                    'client_secret' => config('services.passport.client_secret'),
                    'username' => $request->username,
                    'password' => $request->password,
                    ]
                    ]);


                    return $response->getBody();
                } catch (\GuzzleHttp\Exception\BadResponseException $e) {
                    if ($e->getCode() === 400) {
                        return response()->json('Invalid Request. Please enter a username or a password.', $e->getCode());
                    } else if ($e->getCode() === 401) {
                        return response()->json('Your credentials are incorrect. Please try again', $e->getCode());
            }
            return response()->json('Something went wrong on the server.', $e->getCode());
        }
    }

我想做这样的事情

$user = User::where('email', $request->username)->get();

$token = $response->getBody();

return response()->json($token, $user);

但是，当我尝试此操作时，出现错误。任何帮助将不胜感激。当前，我必须在访问令牌之后单独发出请求以返回用户数据，因此我希望它同时发生。...

Answer 1

json方法的第一个参数可以是一个数组：

return response()->json(['token' => $token, 'user' => $user ]);

Answer 2

我被推荐以下discussion，它解决了我的问题。这是最终代码

public function login(Request $request)
    {
        $http = new \GuzzleHttp\Client;

        try {
            $response = $http->post(config('services.passport.login_endpoint'), [
                'form_params' => [
                    'grant_type' => 'password',
                    'client_id' => config('services.passport.client_id'),
                    'client_secret' => config('services.passport.client_secret'),
                    'username' => $request->username,
                    'password' => $request->password,
                    ]
                    ]);

                    $token = $response->getBody();

                    $data = json_decode($token, true);

                    $user = User::where('email', $request->username)->with('role.rules')->get();

                    $rules = $user->pluck('role')->collapse()->pluck('rules');

                    $rules->put('access_token', $data['access_token']);

                    return response()->json($rules);
                } catch (\GuzzleHttp\Exception\BadResponseException $e) {
                    if ($e->getCode() === 400) {
                        return response()->json('Invalid Request. Please enter a username or a password.', $e->getCode());
                    } else if ($e->getCode() === 401) {
                        return response()->json('Your credentials are incorrect. Please try again', $e->getCode());
            }
            return response()->json('Something went wrong on the server.', $e->getCode());
        }
    }

有没有办法在Laravel中包含带有Http / Guzzle响应的数据

2 个答案: