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scripts/get_release_branch.py
53 строки
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Ken McGrady
Update copyright header to 2026 (#13491)
02 янв 2026, 16:21
Не верифицирован
02 янв 2026, 16:21
374540a
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#!/usr/bin/env python # Copyright (c) Streamlit Inc. (2018-2022) Snowflake Inc. (2022-2026) # # 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. """Retrieve the branch name from the release PR.""" from __future__ import annotations from typing import Any, cast import requests # Assumes there is only one open pull request with a release/ branch def check_for_release_pr(pull: dict[str, Any]) -> str | None: label = pull["head"]["label"] if label.find("release/") != -1: return cast("str", pull["head"]["ref"]) return None def get_release_branch() -> str | None: """Retrieve the release branch from the release PR.""" url = "https://api.github.com/repos/streamlit/streamlit/pulls" response = requests.get(url).json() # Response is in an array, must map over each pull (dict) for pull in response: ref = check_for_release_pr(pull) if ref is not None: return ref return None def main() -> None: print(get_release_branch()) if __name__ == "__main__": main()