如何创建使用其他规则的runfiles路径的自定义bazel构建规则？

Question

我正在尝试创建一个自定义构建规则来从py_binary输出构建一个pip包.要创建我的pip包,我想调用一个shell脚本.shell脚本通过从py_binary输出的runfiles创建一个zip文件来构建一个pip包.

例如,假设我有

py_binary(
    name = "some_binary",
    srcs = ["some_binary.py"],
    srcs_version = "PY2AND3",
)

建立这个规则会产生

bazel-bin/some_binary.runfiles

我现在想创建一个自定义构建规则,它将使用位置bazel-bin/some_binary.runfiles调用我的shell脚本

我尝试创建一个宏

def build_pip_package(
    name, py_binary=None, setup_file=None):
  """Create a pip package from a py_binary.

  The source file should be a text file with python formatting i.e.

  Args:
    name: Name for the rule.
    py_binary: Build rule producing the py_binary
    setup_file: Build rule producing the setup.py file to use to produce
      the package.
  """
  output= "somefile"
  native.genrule(
      name=name,
      outs=[output],
      cmd="echo $(location //:build_pip_package_script) "
      + "--py_runfiles_path=$(locations %s)" % py_binary
      + " --setup_file=$(location %s) " % setup_file,
      tools=["//:build_pip_package_script"],
      srcs=[setup_file, py_binary])

这最终会调用我的shell脚本

bazel-out/local-fastbuild/bin/some_binary/model_train some_binary.py 

如何使用与some_binary目标对应的runfiles目录的位置来调用我的脚本.

Answer 1

如果将py_binary放在tools属性而不是srcs中,bazel将包含runfiles树.你可以通过它访问它"$(location %s).runfiles" % py_binary