Question

我正在尝试在我的项目中将Laravel从版本5.1更新到5.2，我已经从文档中跟踪了这个upgrade guide，但是现在我在验证失败时得到了这个HttpResponseException

 * Handle a failed validation attempt.
 *
 * @param  \Illuminate\Contracts\Validation\Validator  $validator
 * @return mixed
 *
 * @throws \Illuminate\Http\Exception\HttpResponseException
 */
protected function failedValidation(Validator $validator)
{
    throw new HttpResponseException($this->response(
        $this->formatErrors($validator)
    ));
}

在5.1中，框架会自动重定向到上一个URL并显示验证错误。

这是我的验证请求

namespace domain\funcao\formRequest;

use autodoc\Http\Requests\Request;

class StoreFuncaoRequest extends Request
{
    /**
     * 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 [
            'codigo' => 'required|max:255|unique:funcao,codigo,'.$this->input('id').',id,deleted_at,NULL',
            'nome' => 'required|max:255|unique:funcao,nome,'.$this->input('id').',id,deleted_at,NULL'
        ];
    }
}

我已根据指南

更新了我的异常处理程序

class Handler extends ExceptionHandler
{
    /**
     * A list of the exception types that should not be reported.
     *
     * @var array
     */
    protected $dontReport = [
        \Illuminate\Auth\Access\AuthorizationException\AuthorizationException::class,
        \Illuminate\Database\Eloquent\ModelNotFoundException::class,
        \Illuminate\Foundation\ValidationException\ValidationException::class,
        \Symfony\Component\HttpKernel\Exception\HttpException::class,
    ];
    ...
}

有人有这个问题吗？

Answer 1

我找到了此错误的原因，我在我的Exception Handler类中手动从Whoops调用了PrettyPageHandler。

class Handler extends ExceptionHandler
{
    /**
     * A list of the exception types that should not be reported.
     *
     * @var array
     */
    protected $dontReport = [
        \Illuminate\Auth\Access\AuthorizationException\AuthorizationException::class,
        \Illuminate\Database\Eloquent\ModelNotFoundException::class,
        \Illuminate\Foundation\ValidationException\ValidationException::class,
        \Symfony\Component\HttpKernel\Exception\HttpException::class,
    ];

    ...

    /**
     * Render an exception into an HTTP response.
     *
     * @param  \Illuminate\Http\Request  $request
     * @param  \Exception  $e
     * @return \Illuminate\Http\Response
     */
    public function render($request, Exception $e)
    {
        // I just needed to remove this call to get rid of the problem
        if (config('app.debug'))
        {
            return $this->renderExceptionWithWhoops($e);
        }

        return parent::render($request, $e);
    }

    /**
     * Render an exception using Whoops.
     * 
     * @param  \Exception $e
     * @return \Illuminate\Http\Response
     */
    protected function renderExceptionWithWhoops(Exception $e)
    {
        $whoops = new \Whoops\Run;
        $whoops->pushHandler(new \Whoops\Handler\PrettyPageHandler());

        return new \Illuminate\Http\Response(
            $whoops->handleException($e),
            $e->getStatusCode(),
            $e->getHeaders()
        );
    }    
}

我仍在使用Whoops，但现在自动通过Laravel Exceptions

Answer 2

这是我遇到这个问题时遇到的第一个问题，虽然对我而言Sentry不是哎呀。

以下是一些例外以及我如何处理它们：

public function render($request, Exception $e)
{
    if ($e instanceof TokenMismatchException) {
        return redirect()->back()->withInput()->with('error', 'Your Session has Expired');
    }
    if ($e instanceof HttpResponseException) {
        return $e->getResponse();
    }
    return response()->view('errors.500', [
        'sentryID' => $this->sentryID,
    ], 500);
}

我通过简单地以HttpResponseException类的方式返回响应来处理ExceptionHandler。

Answer 3

您所要做的就是，只需在您的自定义FormRequest类内的protected failedValidation（）中编写业务逻辑，如下所示

use Illuminate\Http\Exceptions\HttpResponseException;
use Illuminate\Contracts\Validation\Validator;

/**
* [failedValidation [Overriding the event validator for custom error response]]
* @param  Validator $validator [description]
* @return [object][object of various validation errors]
*/
public function failedValidation(Validator $validator) { 
     // write your business logic here otherwise it will give same old JSON response
    throw new HttpResponseException(response()->json($validator->errors(), 422)); 
}

Laravel请求验证在5.2版中抛出HttpResponseException

3 个答案: