[英]How to access the $container within middleware class in Slim v3?
I've been reading that in Slim v2, $app was bound to the middleware class. 我一直在读Slim v2,$ app被绑定到中间件类。 I'm finding this not to be the case in v3?
我发现v3不是这样的吗? Below is my middleware class, but I'm just getting undefined:
下面是我的中间件类,但我只是未定义:
<?php
namespace CrSrc\Middleware;
class Auth
{
/**
* Example middleware invokable class
*
* @param \Psr\Http\Message\ServerRequestInterface $request PSR7 request
* @param \Psr\Http\Message\ResponseInterface $response PSR7 response
* @param callable $next Next middleware
*
* @return \Psr\Http\Message\ResponseInterface
*/
public function __invoke($request, $response, $next)
{
// before
var_dump($this->getContainer()); // method undefined
var_dump($this->auth); exit; // method undefined
if (! $this->get('auth')->isAuthenticated()) {
// Not authenticated and must be authenticated to access this resource
return $response->withStatus(401);
}
// pass onto the next callable
$response = $next($request, $response);
// after
return $response;
}
}
What's the correct way to access the DI container within middleware? 在中间件中访问DI容器的正确方法是什么? I'm guessing there ought to be a way?
我猜应该有办法吗?
A bit late to the party but might help others... You have to inject your container when instantiating the middleware 派对有点晚,但可能会帮助其他人......在实例化中间件时,你必须注入容器
$container = $app->getContainer();
$app->add(new Auth($container));
And your middleware needs a constructor 而你的中间件需要一个构造函数
<?php
namespace CrSrc\Middleware;
class Auth
{
private $container;
public function __construct($container) {
$this->container = $container
}
/**
* Example middleware invokable class
*
* @param \Psr\Http\Message\ServerRequestInterface $request PSR7 request
* @param \Psr\Http\Message\ResponseInterface $response PSR7 response
* @param callable $next Next middleware
*
* @return \Psr\Http\Message\ResponseInterface
*/
public function __invoke($request, $response, $next)
{
// $this->container has the DI
}
}
LE: Expanding a bit the initial answer, the container gets injected in the constructor if you supply the middleware as a class string LE:扩展一点初始答案,如果您将中间件作为类字符串提供,则容器将被注入构造函数中
$app->add('Auth');
or 要么
$app->add('Auth:similarToInvokeMethod')
As far as I understand the code, Slim (v3) works the following way: 据我所知,代码,Slim(v3)的工作方式如下:
bindTo
with container as param. bindTo
与容器作为param。 if you pass a class/string that resolves to a class, then Slim creates the instance and passes the container as param to the constructor 如果你传递一个解析为类的类/字符串,那么Slim创建实例并将容器作为param传递给构造函数
<?php $app->add(Auth);
otherwise (eg if you add a middleware instance created earlier) then it looks like you have to take care of passing all necessary references. 否则(例如,如果你添加一个先前创建的中间件实例),那么看起来你必须要处理所有必要的引用。
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