# Copyright 2023 The Bazel Authors. All rights reserved.
#
# Licensed under the Apache License, Version 2.0 (the "License");
# you may not use this file except in compliance with the License.
# You may obtain a copy of the License at
#
#     http://www.apache.org/licenses/LICENSE-2.0
#
# Unless required by applicable law or agreed to in writing, software
# distributed under the License is distributed on an "AS IS" BASIS,
# WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
# See the License for the specific language governing permissions and
# limitations under the License.

"""
Normalize a PyPI package name to allow consistent label names

Note we chose `_` instead of `-` as a separator as there are certain
requirements around Bazel labels that we need to consider.

From the Bazel docs:
> Package names must be composed entirely of characters drawn from the set
> A-Z, a–z, 0–9, '/', '-', '.', and '_', and cannot start with a slash.

However, due to restrictions on Bazel labels we also cannot allow hyphens.
See https://github.com/bazelbuild/bazel/issues/6841

Further, rules_python automatically adds the repository root to the
PYTHONPATH, meaning a package that has the same name as a module is picked
up. We workaround this by prefixing with `<hub_name>_`.

Alternatively we could require
`--noexperimental_python_import_all_repositories` be set, however this
breaks rules_docker.
See: https://github.com/bazelbuild/bazel/issues/2636

Also see Python spec on normalizing package names:
https://packaging.python.org/en/latest/specifications/name-normalization/
"""

def normalize_name(name):
    """normalize a PyPI package name and return a valid bazel label.

    Args:
        name: str, the PyPI package name.

    Returns:
        a normalized name as a string.
    """
    name = name.replace("-", "_").replace(".", "_").lower()
    if "__" not in name:
        return name

    # Handle the edge-case where there are consecutive `-`, `_` or `.` characters,
    # which is a valid Python package name.
    return "_".join([
        part
        for part in name.split("_")
        if part
    ])