相对进口的噩梦,pep 366如何运作？

Question

我有一个"规范的文件结构"(我给出了明智的名称来简化阅读):

mainpack/

  __main__.py
  __init__.py 

  - helpers/
     __init__.py
     path.py

  - network/
     __init__.py
     clientlib.py
     server.py

  - gui/
     __init__.py
     mainwindow.py
     controllers.py

在此结构中,例如,每个包中包含的模块可能希望helpers通过相对导入来访问实用程序,例如:

# network/clientlib.py
from ..helpers.path import create_dir

该程序__main__.py以这种方式使用该文件"作为脚本"运行:

python mainpack/

试图按照PEP 366我已经把__main__.py这些行:

___package___ = "mainpack"
from .network.clientlib import helloclient 

但是在跑步时:

$ python mainpack 
Traceback (most recent call last):
  File "/usr/lib/python2.6/runpy.py", line 122, in _run_module_as_main
    "__main__", fname, loader, pkg_name)
  File "/usr/lib/python2.6/runpy.py", line 34, in _run_code
    exec code in run_globals
  File "path/mainpack/__main__.py", line 2, in <module>
    from .network.clientlib import helloclient
SystemError: Parent module 'mainpack' not loaded, cannot perform relative import

怎么了？处理和有效使用相对进口的正确方法是什么？

我也尝试将当前目录添加到PYTHONPATH,没有任何变化.

Answer 1

PEP 366中给出的"样板" 似乎不完整.虽然它设置__package__变量,但它实际上并不导入包,这也是允许相对导入工作所必需的. extraneon的解决方案正在走上正轨.

请注意,仅仅包含包含模块的目录是不够的,sys.path需要显式导入相应的包.以下似乎是比PEP 366中给出的更好的样板,用于确保无论如何调用python模块(通过常规import,或使用python -m,或python来自任何位置),都可以执行python模块:

# boilerplate to allow running as script directly
if __name__ == "__main__" and __package__ is None:
    import sys, os
    # The following assumes the script is in the top level of the package
    # directory.  We use dirname() to help get the parent directory to add to
    # sys.path, so that we can import the current package.  This is necessary 
    # since when invoked directly, the 'current' package is not automatically
    # imported.
    parent_dir = os.path.dirname(os.path.dirname(os.path.abspath(__file__)))
    sys.path.insert(1, parent_dir)
    import mypackage
    __package__ = str("mypackage")
    del sys, os

# now you can use relative imports here that will work regardless of how this
# python file was accessed (either through 'import', through 'python -m', or 
# directly.

如果脚本不在包目录的顶层,并且您需要在顶层下面导入模块,则os.path.dirname必须重复该脚本,直到parent_dir包含顶层的目录为止.

Answer 2

加载代码似乎是这样的:

    try:
        return sys.modules[pkgname]
    except KeyError:
        if level < 1:
            warn("Parent module '%s' not found while handling "
                 "absolute import" % pkgname, RuntimeWarning, 1)
            return None
        else:
            raise SystemError, ("Parent module '%s' not loaded, cannot "
                                "perform relative import" % pkgname)

这让我觉得你的模块可能不在sys.path上.如果您启动Python(通常)并在提示符下键入"import mainpack",它会做什么？它应该能够找到它.

我自己尝试过并得到了同样的错误.读了一下后,我找到了以下解决方案:

# foo/__main__.py
import sys
mod = __import__('foo')
sys.modules["foo"]=mod

__package__='foo'
from .bar import hello

hello()

这对我来说似乎有些神圣,但确实有效.诀窍似乎是确保foo加载包,因此导入可以是相对的.

Answer 3

灵感来自extraneon和taherh的答案,这里有一些代码运行文件树,直到文件用完为止,__init__.py以构建完整的软件包名称.这绝对是hacky,但无论目录树中文件的深度如何,它似乎都能正常工作.这似乎是绝对进口大量的鼓励.

import os, sys
if __name__ == "__main__" and __package__ is None:
    d,f = os.path.split(os.path.abspath(__file__))
    f = os.path.splitext(f)[0]
    __package__ = [f] #__package__ will be a reversed list of package name parts
    while os.path.exists(os.path.join(d,'__init__.py')): #go up until we run out of __init__.py files
        d,name = os.path.split(d) #pull of a lowest level directory name 
        __package__.append(name)  #add it to the package parts list
    __package__ = ".".join(reversed(__package__)) #create the full package name
    mod = __import__(__package__) #this assumes the top level package is in your $PYTHONPATH
    sys.modules[__package__] = mod  #add to modules