[英]How to access static property of Laravel Controller in PhpSpec?
I am trying to write a class which will read the routes given by the constructor it's constructed with. 我正在尝试编写一个类,它将读取由它构造的构造函数给出的路由。 However, even after 1h of googling I didn't find anything how to access the
getRouter
method of the Laravel Controller - because it's a static function . 然而,即使在谷歌搜索1小时后,我也没有找到任何方法来访问Laravel控制器的
getRouter
方法 - 因为它是一个静态函数 。 I've tried many things but most of the time I got following error: 我尝试了很多东西,但大部分时间我都遇到了以下错误:
Uncaught Error: Using $this when not in object context in
未捕获错误:在不在对象上下文中时使用$ this
vendor/phpspec/prophecy/src/Prophecy/Doubler/Generator/ClassCreator.php(49) :vendor / phpspec / prophecy / src / Prophecy / Doubler / Generator / ClassCreator.php(49):
eval()'d code:13eval()'代码:13
Stack trace: #0 [internal function]: Double\\Illuminate\\Routing\\Controller\\P4::getRouter()堆栈跟踪:#0 [内部功能]:Double \\ Illuminate \\ Routing \\ Controller \\ P4 :: getRouter()
How can I achieve this or is this just not possible with PhpSpec? 我怎样才能实现这一目标,或者这对PhpSpec来说是不可能的?
My Spec: 我的规格:
use Illuminate\Routing\Controller;
use PhpSpec\ObjectBehavior;
use Prophecy\Argument;
class OptionDescriberSpec extends ObjectBehavior
{
function let(Controller $controller)
{
$this->beConstructedWith($controller);
}
function it_should_read_the_aviable_routes_of_the_controller()
{
$this->getController()->getRouter()->shouldReturn('Router');
$this->render()->shouldReturn('Router');
}
}
My Class: 我的课:
use Illuminate\Routing\Controller;
class OptionDescriber
{
/**
* @var Controller
*/
protected $controller;
/**
* OptionDescriber constructor.
*
* @param Controller $controller
*/
public function __construct(Controller $controller)
{
$this->controller = $controller;
}
public function render()
{
return $this->controller->getRouter();
}
}
function it_should_read_the_aviable_routes_of_the_controller(Controller $controller, Router $router)
{
$this->getController()->willReturn($controller);
$controller->getRouter()->willReturn($router);
$this->render()->shouldReturn($router);
}
Alltough it's not recommended to spec controllers. 尽管不建议使用规格控制器。 Make the controller as thin as possible and move the 'meat' into a service that is used by the controller.
使控制器尽可能薄,并将“肉”移动到控制器使用的服务中。
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