package_resources.py 文件源码

def get_package_and_resource_name(path):
    """
    This method will return the package name and resource name from a path.

    Arguments:
    path    Path to parse for package and resource name.
    """
    package = None
    resource = None
    path = _normalize_to_sublime_path(path)
    if os.path.isabs(path):
        packages_path = _normalize_to_sublime_path(sublime.packages_path())
        if path.startswith(packages_path):
            package, resource = _search_for_package_and_resource(path, packages_path)

        if int(sublime.version()) >= 3006:
            packages_path = _normalize_to_sublime_path(sublime.installed_packages_path())
            if path.startswith(packages_path):
                package, resource = _search_for_package_and_resource(path, packages_path)

            packages_path = _normalize_to_sublime_path(os.path.dirname(sublime.executable_path()) + os.sep + "Packages")
            if path.startswith(packages_path):
                package, resource = _search_for_package_and_resource(path, packages_path)
    else:
        path = re.sub(r"^Packages/", "", path)
        split = re.split(r"/", path, 1)
        package = split[0]
        package = package.replace(".sublime-package", "")
        resource = split[1]

    return (package, resource)