bazel_namespace_package_hack.py
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1 # Copyright 2019 The gRPC Authors
2 #
3 # Licensed under the Apache License, Version 2.0 (the "License");
4 # you may not use this file except in compliance with the License.
5 # You may obtain a copy of the License at
6 #
7 # http://www.apache.org/licenses/LICENSE-2.0
8 #
9 # Unless required by applicable law or agreed to in writing, software
10 # distributed under the License is distributed on an "AS IS" BASIS,
11 # WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
12 # See the License for the specific language governing permissions and
13 # limitations under the License.
14 
15 import os
16 import site
17 import sys
18 
19 _GRPC_BAZEL_RUNTIME_ENV = "GRPC_BAZEL_RUNTIME"
20 
21 
22 # TODO(https://github.com/bazelbuild/bazel/issues/6844) Bazel failed to
23 # interpret namespace packages correctly. This monkey patch will force the
24 # Python process to parse the .pth file in the sys.path to resolve namespace
25 # package in the right place.
26 # Analysis in depth: https://github.com/bazelbuild/rules_python/issues/55
28  """Add valid sys.path item to site directory to parse the .pth files."""
29  # Only run within our Bazel environment
30  if not os.environ.get(_GRPC_BAZEL_RUNTIME_ENV):
31  return
32  items = []
33  for item in sys.path:
34  if os.path.exists(item):
35  # The only difference between sys.path and site-directory is
36  # whether the .pth file will be parsed or not. A site-directory
37  # will always exist in sys.path, but not another way around.
38  items.append(item)
39  for item in items:
40  site.addsitedir(item)
tests.bazel_namespace_package_hack.sys_path_to_site_dir_hack
def sys_path_to_site_dir_hack()
Definition: bazel_namespace_package_hack.py:27


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autogenerated on Thu Mar 13 2025 02:58:36