Laravel在哪里存储配置的实例逻辑？

Question

I am using Laravel for a Github API project.

I want to set up a generic Guzzle client instance that has been pre-configured. I need this because with about 90% of my Guzzle requests, the Guzzle client will need an access_token header and a base_uri.

// Return a new configured guzzle client.
return new Client([
    'base_uri' => 'https://api.github.com',
    'headers' => [
        'Authorization' => "token {$access_token}"
    ]
]);

I know I can bind this in the container I am currently doing it like so inside a custom middleware:

if (Auth::check()) {
    $this->app->singleton(Client::class, function() {
        // Grab the current user.
        $user = Auth::user();

        $access_token = decrypt($user->access_token);

        // Return a new configured Guzzle instance.
        return new Client([
            'base_uri' => 'https://api.github.com',
            'headers' => [
                'Authorization' => "token {$access_token}"
            ]
        ]);
    });
}

I am simply overriding the default Guzzle client with my pre-configured concrete instance. This works fine but the problem I am running into is that I can not do the following in a controller:

public function __construct(Container $container)
{
    $guzzle = $container->make(Client::class);
}

If I use the above I will simply get the default implementation of Guzzle and not my pre-configured instance. I am guessing this it because the controller's constructor runs before the middleware and is it simply not set yet.

I still want to share the property to other methods in the controller though. How could I fix this?

Answer 1

I would create a new class for api interactions. In the constructor, initiate your guzzle object and then create methods for various api calls that use that guzzle object.