Laravel 5、检查类是否在容器中注册

Laravel 5, check if class is registered in the container

提问by Silvio Sosio

Is there any way to check if a class exists in Laravel 5?

有没有办法检查 Laravel 5 中是否存在类？

I had this solution for Laravel 4: try to make a specific class, and if I get a ReflectionException, I use a generic class.
In Laravel 5 looks like I can't intercept the ReflectionExceptionand I get a "Whoops".

我为 Laravel 4 提供了这个解决方案：尝试创建一个特定的类，如果我得到一个ReflectionException，我就使用一个泛型类。
在 Laravel 5 中，我似乎无法拦截ReflectionException并且收到“哎呀”。

I was wondering if there is some better way to do this.

我想知道是否有更好的方法来做到这一点。

try {
    $widgetObject = \App::make($widget_class);
} catch (ReflectionException $e) {
    $widgetObject = \App::make('WidgetController');
    $widgetObject->widget($widget);
}

回答by lukasgeiter

Why don't you just use the PHP function class_exists?

为什么不直接使用 PHP 函数class_exists？

if(class_exists($widget_class)){
    // class exists
}

回答by Aderemi Dayo

Check whether your class is set in bindings by doing

通过执行检查您的类是否设置在绑定中

app()->bound($classname);

回答by Alexander Konotop

\App::bound()might be the proper way.

\App::bound()可能是正确的方法。

Latest laravel versions (Maybe >= 5.3, I don't know exactly) register service providers in alittle different way by default.

默认情况下，最新的 Laravel 版本（也许 >= 5.3，我不知道确切）以稍微不同的方式注册服务提供者。

For example, a new way of registering:

例如，一种新的注册方式：

$this->app->singleton(MyNamespace\MyClass::class, function()
{
    /* do smth */ }
);

instead of the old one:

而不是旧的：

$this->app->singleton('MyClassOrAnyConvenientName', function()
{
    /* do smth */ }
);

As a result we should use App::make('\MyNamespace\MyClass')instead of App::make('MyClassOrAnyConvenientName')to resolve a service.

因此，我们应该使用App::make('\MyNamespace\MyClass')而不是App::make('MyClassOrAnyConvenientName')来解析服务。

We maintain a library which has to support both versions. So we use \App::bound()to determine whether old or new format of service name is registered in container. class_exists()did actually work for newer laravel but didn't work as expected for older ones because in old systems we didn't have a properly named Facadefor that service (Facade's name differed from registered service name) and class_exists()returned false.

我们维护一个必须支持两个版本的库。所以我们\App::bound()用来判断容器中注册的是旧格式还是新格式的服务名。class_exists()实际上确实适用于较新的 Laravel，但对于较旧的 Laravel 没有按预期工作，因为在旧系统中，我们没有Facade为该服务正确命名（Facade 的名称与注册的服务名称不同）并class_exists()返回false.

回答by Jamie Holly

Use the getProvider method on the Application container:

在 Application 容器上使用 getProvider 方法：

if (app()->getProvider($classname)) {
   // Do what you want when it exists.
 }

It's been available since 5.0 and can be viewed here.

它从 5.0 开始可用，可以在这里查看。