一、錯誤提示php
FatalErrorException in Handler.php line 38: Uncaught TypeError: Argument 1 passed to App\Exceptions\Handler::report() must be an instance of Exception, instance of Error given, called in D:\www\activity\vendor\compiled.php on line 1817 and defined in D:\www\activity\app\Exceptions\Handler.php:38 Stack trace: #0 D:\www\activity\vendor\compiled.php(1817): App\Exceptions\Handler->report(Object(Error)) #1 [internal function]: Illuminate\Foundation\Bootstrap\HandleExceptions->handleException(Object(Error)) #2 {main} thrown
緣由:D:wwwactivityvendorcompiled.php on line 1817 的變量$e不是Exception的實例對象(對錯誤提示的翻譯……^.^笑cry)app
二、解決方案
在提示的錯誤地方加上變量$e的實例判斷,若是不是Exception類型,就new一個this
if (!$e instanceof \Exception) { $e = new FatalThrowableError($e); }
new完以後的樣子:翻譯
public function handleException($e) { if (!$e instanceof \Exception) { $e = new FatalThrowableError($e); } $this->getExceptionHandler()->report($e); if ($this->app->runningInConsole()) { $this->renderForConsole($e); } else { $this->renderHttpResponse($e); } }