fix: use full url for Swagger Ui bundles urls

[nathanaelhoun]

Hello! I wanted to use your package on an application hosted in a subdirectory (like https://example.net/my-laravel-application/), but it failed to load the Swagger bundle because the link is generated from the root of the application.

This proposed change fixes this behaviour by using the url() helper to generate a fully qualified link to the openapi file.

I failed to add a unit test 😭

I tried to add a unit test to test for this behaviour, but I couldn't get it to work correctly.

• By default, the url() helper doesn't use the config from app.url in the CLI (as far as I can say)
• So I added URL::forceRootUrl(config('app.url'));, but then the request gives a 404 as the test request now includes the subdir in it's path
• The last solution was to override the prepareUrlForRequest() method to remove the url() helper call from it, but I'm not keen to override a framework method. It works though.

    /** @test */
    public function it_supports_application_sub_dir_and_swagger_sub_path()
    {
        config()->set('app.url', 'http://localhost/application-subdir/');

        // https://github.com/laravel/framework/issues/18613#issuecomment-342033733
        URL::forceRootUrl(config('app.url'));

        $this->assertEquals('http://localhost/application-subdir/path/with/multiple/segments/swagger-with-versions', url('path/with/multiple/segments/swagger-with-versions'));

        $this->get('/path/with/multiple/segments/swagger-with-versions')
            ->assertStatus(200)
            ->assertSee('url: \'http://localhost/application-subdir/path/with/multiple/segments/swagger-with-versions/v1\'', false)
            ->assertSee('url: \'http://localhost/application-subdir/path/with/multiple/segments/swagger-with-versions/v2\'', false);
    }

    /**
     * @overrides Do not use the `url()` helper, because the router doesn't use the setting from config('app.url')
     */
    protected function prepareUrlForRequest($uri)
    {
        if (str_starts_with($uri, '/')) {
            $uri = substr($uri, 1);
        }

        return trim($uri, '/');
    }

[nathanaelhoun]

Hi @gdebrauwer, is there anything that I can do to help getting this PR reviewed?

[gdebrauwer]

Lost track of this PR 😄

Fixed in v0.9.2