关于数据获取时不存在时的404报错处理方案

在ArticleContronller.php中出现数据获取不存在时

public function show($id){    $article=Arrticle::find($id)    if(is_null($article)){        abort(404)    }    return view(articles.show,compact('article'));}

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注：关于abort()与exit()
exit和abort都是用来终止程序的函数，他们的不同如下：

exit会做一些释放工作：释放所有的静态的全局的对象，缓存，关掉所有的I/O通道，然后终止程序。如果有函数通过atexit来注册，还会调用注册的函数。不过，如果atexit函数扔出异常的话，就会直接调用terminate。

abort：立刻terminate程序，没有任何清理工作。

参考：http://bbs.csdn.net/topics/390130642

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注：
Q:Laravel app:abort(404) not showing 404 page

I have a route of /path/{id}, and I want my controller to detect when the id is invalid and product a 404 error page. I tried using App::abort(404);, but this doesn’t produce a 404 error - just a generic exception. How do I make it generate a 404 page?

A:You have to use the Response::make or Response::view functions to generate a specific page. You can pass the error code as an argument to that. Like this:

Response::make("Page not found", 404);

Or to use a specific view, use:

return Response::view('404', array(), 404);

To display the view app/views/404.php with the http response code 404. Remember to replace the slashes with dots for the view name, so if your file is at app/views/errors/404.php, you will need to use the view name ‘errors.404’ for laravel to find it.

参考：http://stackoverflow.com/questions/22011202/laravel-appabort404-not-showing-404-page

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注：
当命名空间不一致时，注意使用方式

\App::abort(404);app()->abort(404);

参考https://laravel-china.org/topics/1385/laravel-5-how-to-use-app-correctly-abort-404

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使用findorFail()

public function show($id){    $article=Arrticle::findorFail($id)    if(is_null($article)){        abort(404)    }    return view(articles.show,compact('article'));}