Question

我正在尝试拦截laravel中的错误，我发现了一种不错的方法：

模拟错误：

public function index(){       

 $users = User::all(); //<-SQL exeception here             

 return response()->json(['message'=>'ok'], 200);
}

app / Exceptions / Handler.php

public function report(Exception $exception)
{
    dd($exception); //<-intercept my error here
    parent::report($exception);
}

工作得很好，我可以做任何我想做的事，但有错误，但是当我使用try-catch块时，我的拦截器不起作用：

再次模拟错误

public function index(){

 try {

   $users = User::all();//<-SQL exeception here 

 } catch (\Throwable $th) {

   error_log('Error handled');

   //MyInterceptor::manuallyIntercept($th);

 }        

    return response()->json(['message'=>'ok'], 200);
}

是否有一种干净的方法来以编程方式拦截所有已处理的错误？

Answer 1

不是报告方法，您需要在List<WebElement> rows = StockWatchlistElements.lnkaStockWatchlist(driver); int count = rows.size(); System.out.println("ROW COUNT : " + count);上使用渲染方法您将看到Handler.php，它将仅返回JSON响应。我只想展示主要思想。

$this->errorResponse

错误响应方法

public function render($request, Exception $exception)
{
    if ($exception instanceof ValidationException) {
        return $this->convertValidationExceptionToResponse($exception, $request);
    }
    if ($exception instanceof ModelNotFoundException) {
        $modelName = strtolower(class_basename($exception->getModel()));
        return $this->errorResponse("Does not exists any {$modelName} with the specified identificator", 404);
    }
    if ($exception instanceof AuthenticationException) {
        return $this->unauthenticated($request, $exception);
    }
    if ($exception instanceof AuthorizationException) {
        return $this->errorResponse($exception->getMessage(), 403);
    }
    if ($exception instanceof MethodNotAllowedHttpException) {
        return $this->errorResponse('The specified method for the request is invalid', 405);
    }
    if ($exception instanceof NotFoundHttpException) {
        return $this->errorResponse('The specified URL cannot be found', 404);
    }
    if ($exception instanceof HttpException) {
        return $this->errorResponse($exception->getMessage(), $exception->getStatusCode());
    }
    if ($exception instanceof QueryException) {
        $errorCode = $exception->errorInfo[1];
        if ($errorCode == 1451) {
            return $this->errorResponse('Cannot remove this resource permanently. It is related with any other resource', 409);
        }
    }
    if (config('app.debug')) {
        return parent::render($request, $exception);
    }
    return $this->errorResponse('Unexpected Exception. Try later', 500);
}

有没有一种干净的方法来拦截Laravel中的* handled *错误？

1 个答案: