python - 将命令添加到 distutils 构建的挂钩？

Question

我在 setup.py 脚本中添加了一个自定义 distutils 命令：

from distutils.command.build_py import build_py

cmdclass = {}
cmdclass['build_qt'] = BuildQt
cmdclass['build_py'] = build_py

setup(..., cmdclass=cmdclass, ...)

有没有办法让它在运行时::

python setup.py build

这首先调用

python setup.py build_qt

自动地？

Answer 1

您可以覆盖build：

from distutils.command.build import build

class my_build(build):
    def run(self):
        self.run_command("build_qt")
        build.run(self)

cmdclass['build'] = my_build

Answer 2

为了添加您自己的命令，您可以将默认的build-command 子类化并扩展其子命令：

class _build(build):
     sub_commands = [('build_qt', None)] + build.sub_commands

...
setup(..., cmdclass={'build': _build, ...})

文档（distutils.cmd.Command）：

# 'sub_commands' formalizes the notion of a "family" of commands,
# eg. "install" as the parent with sub-commands "install_lib",
# "install_headers", etc.  The parent of a family of commands
# defines 'sub_commands' as a class attribute; it's a list of
#    (command_name : string, predicate : unbound_method | string | None)
# tuples, where 'predicate' is a method of the parent command that
# determines whether the corresponding command is applicable in the
# current situation.  (Eg. we "install_headers" is only applicable if
# we have any C header files to install.)  If 'predicate' is None,
# that command is always applicable.
#
# 'sub_commands' is usually defined at the *end* of a class, because
# predicates can be unbound methods, so they must already have been
# defined.  The canonical example is the "install" command.
sub_commands = []