从Laravel中的验证器发送自定义响应

时间:2018-03-17 08:45:14

标签: php laravel laravel-5 laravel-5.2 laravel-5.3

我有一个注册用户路线,需要nameemailpassword。如果数据正确,即存在唯一的电子邮件和参数,则它可以正常工作,但如果用户已经注册,则Laravel会以自己的格式发送自动错误消息。我希望返回格式在成功或失败的情况下保持一致。

成功注册返回数据:

{
    "status": "success",
    "token": "eyJ0eXAiOiJKV1QiLCJhbGciOiJIUzI1NiJ9.eyJzdWIiOjUsImlzcyI6Imh0dHA6Ly8xMjcuMC4wLjE6ODAwMC9hcGkvYXV0aC9yZWdpc3RlciIsImlhdCI6MTUyMTI3NTc5MiwiZXhwIjoxNTIxMjc5MzkyLCJuYmYiOjE1MjEyNzU3OTIsImp0aSI6Ik1wSzJSYmZYU1dobU5UR0gifQ.fdajaDooBTwP-GRlFmAu1gtC7_3U4ygD1TSBIqdPHf0"
}

但如果出现错误,它会以其他格式发送数据。

{"message":"The given data was invalid.","errors":{"email":["The email has already been taken."]}}

我希望他们两个都保持一致。成功返回数据很好。但我想在发生故障时自定义数据。像这样:

{"status":"error","message":"The given data was invalid.","errors":{"email":["The email has already been taken."]}}

基本上,我需要status param随每个回复而来。

另外,我在使用Postman时有一个查询,输出是纯HTML,当发生错误时HTML页面是默认的Laravel页面,另一方面当angular发送相同的请求时,错误是json格式,我刚刚粘贴在上面。 因为角度在任何情况下都得到JSON,所以对我来说这很好。但是为什么邮递员没有向我展示这种反应。

注册控制器:

public function register(RegisterRequest $request)
    {
        $newUser = $this->user->create([
            'name' => $request->get('name'),
            'email' => $request->get('email'),
            'password' => bcrypt($request->get('password'))
        ]);
        if (!$newUser) {
            return response()->json(['status'=>'error','message'=>'failed_to_create_new_user'], 500);
        }

        return response()->json([
            'status' => 'success',
            'token' => $this->jwtauth->fromUser($newUser)
        ]);
    }

注册请求处理程序:

 public function authorize()
    {
        return true;
    }

    /**
     * Get the validation rules that apply to the request.
     *
     * @return array
     */
    public function rules()
    {
        return [
            'name' => 'required',
            'email' => 'required | email | unique:users,email',
            'password' => 'required'
        ];
    }

3 个答案:

答案 0 :(得分:1)

如果我理解正确,您总是会在没有“状态”的情况下得到错误回复。键。

您当前的代码会发生什么,有几件事情:

  • RegisterController @ register(RegisterRequest $ request)由路由
  • 调用
  • Laravel认为您使用RegisterRequest类作为参数,并将为您实例化此类。
  • 实例化此类意味着它将直接验证规则。
  • 如果不符合规则,laravel会直接回应发现的错误。
  • 此响应将始终采用laravel的默认布局'并且代码停在那里。

结论:当您的验证规则不符合时,您的代码甚至无法触发。

我已经研究过一个解决方案并提出了这个问题:

public function register(Illuminate\Http\Request $request)
{
    //Define your validation rules here.
    $rules = [
        'name' => 'required',
        'email' => 'required | email | unique:users,email',
        'password' => 'required'
    ];
    //Create a validator, unlike $this->validate(), this does not automatically redirect on failure, leaving the final control to you :)
    $validated = Illuminate\Support\Facades\Validator::make($request->all(), $rules);

    //Check if the validation failed, return your custom formatted code here.
    if($validated->fails())
    {
        return response()->json(['status' => 'error', 'messages' => 'The given data was invalid.', 'errors' => $validated->errors()]);
    }

    //If not failed, the code will reach here
    $newUser = $this->user->create([
        'name' => $request->get('name'),
        'email' => $request->get('email'),
        'password' => bcrypt($request->get('password'))
    ]);
    //This would be your own error response, not linked to validation
    if (!$newUser) {
        return response()->json(['status'=>'error','message'=>'failed_to_create_new_user'], 500);
    }

    //All went well
    return response()->json([
        'status' => 'success',
        'token' => $this->jwtauth->fromUser($newUser)
    ]);
}

现在,不符合您的验证规则仍会触发错误,但是您的错误,而不是laravel的内置错误:)

我希望它有所帮助!

答案 1 :(得分:1)

在 Laravel 8 中,我添加了我的自定义 invalidJson 并带有 "success": false:

在 app/Exceptions/Handler.php 中:

/**
 * Convert a validation exception into a JSON response.
 *
 * @param  \Illuminate\Http\Request  $request
 * @param  \Illuminate\Validation\ValidationException  $exception
 * @return \Illuminate\Http\JsonResponse
 */
protected function invalidJson($request, ValidationException $exception)
{
    return response()->json([
        'success' => false,
        'message' => $exception->getMessage(),
        'errors' => $exception->errors(),
    ], $exception->status);
}

答案 2 :(得分:0)

这是我想出来的:

function validate(array $rules)
{
    $validator = Validator::make(request()->all(), $rules);
    $errors = (new \Illuminate\Validation\ValidationException($validator))->errors();
    if ($validator->fails()) {
        throw new \Illuminate\Http\Exceptions\HttpResponseException(response()->json(
            [
                'status'     => false,
                'message'    => "Some fields are missing!",
                'error_code' => 1,
                'errors'     => $errors,
            ], \Illuminate\Http\JsonResponse::HTTP_UNPROCESSABLE_ENTITY));
    }
}

创建一个辅助目录 (App\Helpers) 并将其添加到文件中。不要忘记将其添加到您的 composer.json 中

"autoload": {
    "files": [
        "app/Helpers/system.php",
    ],
},

现在您可以在控制器中调用 validate() 并获得您想要的:

validate([
    'email'    => 'required|email',
    'password' => 'required|min:6|max:32',
    'remember' => 'nullable|boolean',
    'captcha'  => 'prod_required|hcaptcha',
]);