Question

我有2个函数使用这样的一个验证器

public function store( Request $request ) {
      $validate = $this->validator( $request );
      if ( $validate->fails() ) {
            return response()->json( [ 'errors' => $validate->errors() ] );
      }
 }

public function update( Request $request, $id ) {
      $validate = $this->validator( $request, $id );

      if ( $validate->fails() ) {
            return response()->json( [ 'errors' => $validate->errors() ] );
      }
}

private function validator( Request $request, $id = "" ) {
      $validator = Validator::make( $request->all(), [
            'name'                  => 'required',
            'email'                 => 'required|email|unique:users,email,' . $id,
            'password'              => 'required|min:6|confirmed',
            'password_confirmation' => "required",
            'role'                  => "required"
      ] );
      if ( $validate->fails() ) {
            return response()->json( [ 'errors' => $validate->errors() ] );
      }
      return $validator;
 }

在这个if验证器中，它返回响应给验证器函数而不是父函数。我想编写函数验证器来检查并给父返回响应，因为我不想检查并一次又一次返回

Answer 1

您可以创建一个继承Request并验证其输入的类。

在控制器方面，您只有：

public function store(SomeRequestValidation $request ) {
      //Do something because it has been validated in SomeRequestValidation
 }

public function update(SomeRequestValidation $request) {
      //Do something because it has been validated in SomeRequestValidation
}

注意：如果是Laravel，那么我们在谈论FormRequests

https://laravel.com/docs/5.6/validation#creating-form-requests

编辑：一个有关重写请求功能可以完成的小示例。

public function authorize(Request $request)
    {
        if (/*something*/) {
            return true;
        }
        return false;
    }

public function forbiddenResponse()
    {
        return json_encode("Oh no you don't");
        //return response()->view('errors.403');
    }

https://laravel.com/api/5.6/Illuminate/Foundation/Http/FormRequest.html

Answer 2

您可以抛出将直接给出响应的错误，现在无需在函数中一次又一次地返回

throw new HttpResponseException(response()->json($error, 422, $headers));

我认为这会对您有所帮助。

PHP函数父级返回

2 个答案: