symfony 3获得基本网址

Question

I want to get top URL in service YAML like this :

app.meal_list_service:
    class: Acme\MealLogBundle\Service\MealListService
    arguments: ['@app.entity_repository_meal',getBaseUrl()"]

I cant find a good solution for this. I tried getting the URL from

'%router.request_context.scheme%://%router.request_context.host%/'

but this return only first part URL.

Answer 1

Edit:

My mistake. I thought you wanted the root project directory. Try defining the project root path in your parameters.yml and pass it in a way similar to passing the %kernel.root_dir%.

// in parameters.yml
my_project_path: https://stackoverflow.com

// in your service.yml
app.meal_list_service:
    class: Acme\MealLogBundle\Service\MealListService
    arguments: 
        - '@app.entity_repository_meal'
        - '%my_project_path%'

Answer 2

You can do:

app.meal_list_service:
    class: Acme\MealLogBundle\Service\MealListService
    arguments: ['@app.entity_repository_meal', '%request_stack%']

And in your service:

$baseUrl = $requestStack->getCurrentRequest()->getSchemeAndHttpHost();

Answer 3

You can always use __ DIR __ and create path from it . Somtimes it's faster than DI and passing parameters to service.

Look at this , i'm using it in trait to create path for image upload . https://gist.github.com/poznet/9d9b0a8b707a527fa56d9bb1fa1979ad