distutils buildにコマンドを追加するためのフック？

Question

私はsetup.pyスクリプトにカスタムのdistutilsコマンドを追加しました：

from distutils.command.build_py import build_py 

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

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

がするよう実行している場合::

python setup.py build 

この最初の呼び出しにそれを作るための方法があります

python setup.py build_qt 

？あなたがbuild上書きすることができ

Answer 1

：独自のコマンドを追加するために

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

を、デフォルトbuild - コマンドをサブクラス化し、そのサブコマンドを拡張することができます。

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

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

ドキュメント（distutils.cmd。コマンド）：

# '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 = []