import requests


def fetch_file_content(resource_url):
    """
    Fetch the XML content of a resource via its URL
    :param resource_url
    :return: the text content of the file, None if the status code was not 200
    """
    response = requests.get(resource_url, headers={'accept': 'application/xml'})
    if response.status_code != 200:
        return None
    return response.text