sqs.changeMessageVisibility according to [this](https://docs.aws.amazon.com/AWSSimpleQueueService/latest/APIReference/API_ChangeMessageVisibility.html#API_ChangeMessageVisibility_Examples) makes a request directly to the Queue's URL rather than via the SQS Endpoint. (I'm already setting SQS endpoint to LOCALSTACK_HOSTNAME - makes no difference for this call)..

In localstack the queue URL host returned is by default set to the value of HOSTNAME_EXTERNAL, which I'm setting to localstack, but calling this method within a lambda, does not work and I get the error:

```
UnknownEndpoint: Inaccessible host: `localstack'. This service may not be available in the `us-east-1' region. 
```

So, how can I make the queue URL accessible from within a lambda using localstack? Effectively I need to make **http://localstack** accessible from within lambda.

I'm using the following cmd to start localstack:

```
LAMBDA_EXECUTOR=docker-reuse DEBUG=1 LAMBDA_REMOTE_DOCKER=0 DATA_DIR=/tmp/localstack/data HOSTNAME_EXTERNAL=localstack localstack start
```
return get_local_region()
def get_local_region():
    global LOCAL_REGION
    if LOCAL_REGION is None:
        session = boto3.session.Session()
        LOCAL_REGION = session.region_name or ''
    return LOCAL_REGION or config.DEFAULT_REGION
def get_local_service_url(service_name_or_port):
    """ Return the local service URL for the given service name or port. """
    if isinstance(service_name_or_port, int):
        return '%s://%s:%s' % (get_service_protocol(), LOCALHOST, service_name_or_port)
    service_name = service_name_or_port
    if service_name == 's3api':
        service_name = 's3'
    elif service_name == 'runtime.sagemaker':
        service_name = 'sagemaker-runtime'
    service_name_upper = service_name.upper().replace('-', '_').replace('.', '_')
    return os.environ['TEST_%s_URL' % service_name_upper]
def is_service_enabled(service_name):
    """ Return whether the service with the given name (e.g., "lambda") is available. """
    try:
        url = get_local_service_url(service_name)
        assert url
        return is_port_open(url, http_path='/', expect_success=False)
    except Exception:
        return False
def connect_to_resource(service_name, env=None, region_name=None, endpoint_url=None, *args, **kwargs):