Question

在Laravel 5.4中，我们创建了一个类，我们所有的验证请求都继承了，因为我们需要自定义响应。

class APIRequest extends FormRequest
{
    /**
     * Determine if the user is authorized to make this request.
     *
     * @return bool
     */
    public function authorize()
    {
        return false;
    }

    /**
     * Response on failure
     * 
     * @param array $errors
     * @return Response
     */
    public function response(array $errors) {
        $response = new ResponseObject();

        $response->code = ResponseObject::BAD_REQUEST;
        $response->status = ResponseObject::FAILED;
        foreach ($errors as $item) {
            array_push($response->messages, $item);
        }
        return Response::json($response);
    }

    /**
     * Get the validation rules that apply to the request.
     *
     * @return array
     */
    public function rules()
    {
        return [
            //
        ];
    }
}

将扩展此范围的示例请求如下所示

class ResultsGetTermsRequest extends APIRequest
{
    /**
     * Determine if the user is authorized to make this request.
     *
     * @return bool
     */
    public function authorize()
    {
        return true;
    }

    /**
     * Get the validation rules that apply to the request.
     *
     * @return array
     */
    public function rules()
    {
        return [
            'school_id' => 'required|integer',
            'student_id' => 'required|integer',
        ];
    }
}

然后我们的失败样本响应将是

{
    "status": "FAILED",
    "code": "400",
    "messages": [
        [
            "The school id field is required."
        ],
        [
            "The student id field is required."
        ]
    ],
    "result": []
}

然而，这对Laravel 5.5来说已经不再适用了。我注意到他们用failedValidation的响应方法替换了。但是，当请求未经过验证时，这不会返回任何响应。如果我取消对print_r的注释，则会返回一些内容。似乎永远不会执行的唯一行是return语句。我错过了什么？

 public function failedValidation(Validator $validator) {

        $errors = (new ValidationException($validator))->errors();
        $response = new ResponseObject();

        $response->code = ResponseObject::BAD_REQUEST;
        $response->status = ResponseObject::FAILED;
        foreach ($errors as $item) {
            array_push($response->messages, $item);
        }
        //print_r($response);
        return Response::json($response);
    }

Answer 1

我想根据laravel upgrade指南，我们应该返回cookie

HttpResponseException

Answer 2

如果你想从FormRequest类中执行此操作，可能是这样的：

protected function buildResponse($validator)
{
    return response->json([
        'code' => ResponseObject::BAD_REQUEST,
        'status' => ResponseObject::FAILED,
        'messages' => $validator->errors()->all(),
    ]);
 }

protected function failedValidation(Validator $validator)
{
    throw (new ValidationException($validator, $this->buildResponse($validator));
}

这会将您正在构建的响应添加到验证异常中。当异常处理程序尝试呈现它时，它将检查是否设置了response，如果是，它将使用您传递的响应，而不是尝试将ValidationException转换为响应本身。

如果你希望“ALL”验证异常最终以这种格式呈现，我可能只是在异常处理程序级别执行此操作，因为异常处理程序已经能够将这些异常转换为Json，因此您可以更改处理程序本身的格式，基本上不必对默认的FormRequest进行任何调整。

Answer 3

如果您使用的是laravel 5+，则可以通过覆盖invalid()文件中的invalidJson()或App/Exceptions/Handler.php方法来轻松实现这一目标

就我而言，我正在开发API，并且api响应应采用特定格式，因此我在Handler.php文件中添加了以下内容。

/**
     * 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([
            'code'    => $exception->status,
            'message' => $exception->getMessage(),
            'errors'  => $this->transformErrors($exception),

        ], $exception->status);
    }

// transform the error messages,
    private function transformErrors(ValidationException $exception)
    {
        $errors = [];

        foreach ($exception->errors() as $field => $message) {
           $errors[] = [
               'field' => $field,
               'message' => $message[0],
           ];
        }

        return $errors;
    }

信用：Origianal Answer

Laravel 5.5验证失败时验证的响应更改格式

3 个答案: