Laravel 不会遵守状态码
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原文地址: http://stackoverflow.com/questions/14704644/
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Laravel won't obey status code
提问by Alex
I just can't understand, and don't know where else to look, as the response status code of the following code is always 200, even if I set it to 400 in the main Response class.
我只是无法理解，也不知道还能在哪里查看，因为以下代码的响应状态代码始终为 200，即使我在主响应类中将其设置为 400。
class Api_Controller extends Base_Controller
{
public function __construct()
{
parent::__construct();
//header("HTTP/1.0 404 Not Found"); ##> This works
//die();
$test = array('1' => '2');
die(Response::json($test, 400));
}
What am I missing? I'm not using any extended class, just the default...
我错过了什么？我没有使用任何扩展类，只是默认...
Update
更新
This is the output of the Response::json...above: http://pastebin.com/RGcinSdg
这是Response::json...上面的输出：http: //pastebin.com/RGcinSdg
As you can see, the output has the values that has been set... but still for some reason, returns 200
如您所见，输出具有已设置的值……但仍然出于某种原因，返回 200
Update2
更新2
The output of var_dump(http_response_code());is always 200
的输出var_dump(http_response_code());总是200
Update3 - Temporary fix
Update3 - 临时修复
I have activated an extended version of Response::jsonand add the following line to it
我已经激活了 的扩展版本Response::json并向其添加了以下行
http_response_code($status);
But I would still much like to know why doesn't it does it, the way it should
但我仍然很想知道它为什么不这样做，它应该采用的方式
回答by Oddman
You can't return responses from controller constructors - it just doesn't fit with the flow of the request lifecycle in Laravel.
你不能从控制器构造函数返回响应——它只是不符合 Laravel 中请求生命周期的流程。
There's two ways to do this. You can either:
有两种方法可以做到这一点。您可以：
a) Setup a response filter that handles whatever functionality it is you're trying to achieve or b) Force a controller ACTION to return a response. This would be done like so:
a) 设置一个响应过滤器来处理您试图实现的任何功能或 b) 强制控制器 ACTION 返回响应。这将像这样完成：
class Api_Controller extends Base_Controller
{
public $restful = true;
public function get_index()
{
return Response::json($test, 400);
}
}
It DOES work - you're just doing it incorrectly :)
它确实有效 - 你只是做错了:)
回答by ecunado
The same problem happens if you forget the returnstatement:
如果您忘记return语句，也会发生同样的问题：
Response::json(array(
'error' => true,
'msg' => 'Bad request'
), 403);
instead of:
代替：
return Response::json(array(
'error' => true,
'msg' => 'Bad request'
), 403);
回答by Jannie Theunissen
Try the response()->json()syntax.
试试response()->json()语法。
So, for example, to flag validation error from a custom FormResquest, you can do this:
因此，例如，要从自定义 FormResquest 中标记验证错误，您可以执行以下操作：
/**
* Get validation response for the request.
*
* @param array $messages
* @return \Symfony\Component\HttpFoundation\Response
*/
public function response(array $messages)
{
return response()->json($messages, 422);
}
回答by user2436802
To keep Google Webmaster Tools happy I detect and use the following in the header:
为了让 Google 网站管理员工具满意，我检测并在标题中使用以下内容：
META NAME="ROBOTS" CONTENT="NOINDEX, NOFOLLOW"
回答by bstrahija
Controller methods always have to return responses. But I don't think you can return a response from the constructor. You would need to use a filter.
控制器方法总是必须返回响应。但我认为您不能从构造函数返回响应。您将需要使用过滤器。
