有没有一种可靠的方法可以从在 Sphinx conf.py 中执行的 Python 函数中获取调用者模块的路径？

Question

我运行正在 Sphinx 中编写一些自定义 Python 代码，需要获取调用方模块的路径。（本质上这是调用者的 __file__ 对象；我需要解释一个相对于此位置的文件名。）

我可以根据 How to use inspect to get the caller's info from callee in Python?, but apparently I need to interpret this filename in the context of the Python startup directory. (Sometimes inspect.stack()[k][1] is an absolute path but sometimes it is a relative path like conf.py; the inspect.stack() function doesn't seem to document this but unutbu claims in a comment 从 inspect.stack() 获取文件名，它是相对于 Python 启动目录的。 )

Sphinx 做了一些无意中的坏事，就像这个评论：

# This file is execfile()d with the current directory set to its
# containing dir.

所以 os.path.abspath(filename) 不起作用，并且

# If extensions (or modules to document with autodoc) are in another directory,
# add these directories to sys.path here. If the directory is relative to the
# documentation root, use os.path.abspath to make it absolute, like shown here.
sys.path.insert(0, os.path.abspath('extensions'))

所以 sys.path[0] 在我的代码到达它时已损坏。

如果sys.path被修改，如何在Python中找到启动目录？

或者有其他方法获取调用模块的路径吗？

如果我运行

for file,line,w1,w2 in traceback.extract_stack():
    sys.stdout.write('  File "{}", line {}, in {}\n'.format(file,line,w1))

我明白了：

File "c:\app\python\anaconda.6.0\Scripts\sphinx-build-script.py", line 5, in <module>
File "c:\app\python\anaconda.6.0\lib\site-packages\Sphinx-1.4.1-py2.7.egg\sphinx\__init__.py", line 51, in main
File "c:\app\python\anaconda.6.0\lib\site-packages\Sphinx-1.4.1-py2.7.egg\sphinx\__init__.py", line 92, in build_main
File "c:\app\python\anaconda.6.0\lib\site-packages\Sphinx-1.4.1-py2.7.egg\sphinx\cmdline.py", line 243, in main
File "c:\app\python\anaconda.6.0\lib\site-packages\Sphinx-1.4.1-py2.7.egg\sphinx\application.py", line 155, in __init__
File "conf.py", line 512, in setup
[more lines elided, the conf.py is the one that matters]

所以问题是我需要找到 conf.py 的路径，但当前目录已被 Sphinx 更改，所以我不能只做 os.path.abspath(caller_filename)

Answer 1

您可以使用 traceback 模块获得您想要的。我在 PyScripter 中编写了这个示例代码：

import traceback,sys

def demo():
   for file,line,w1,w2 in traceback.extract_stack():
       sys.stdout.write('  File "{}", line {}, in {}\n'.format(file,line,w1))

def foo():
    demo()

foo()

在我的 Windows PC 运行 PyScripter 上给出：

  File "C:\Users\dartypc\AppData\Roaming\PyScripter\remserver.py", line 63, in <module>
  File "C:\Users\dartypc\AppData\Roaming\PyScripter\remserver.py", line 60, in main
  File "C:\Program Files\PyScripter\Lib\rpyc.zip\rpyc\utils\server.py", line 227, in start
  File "C:\Program Files\PyScripter\Lib\rpyc.zip\rpyc\utils\server.py", line 139, in accept
  File "C:\Users\dartypc\AppData\Roaming\PyScripter\remserver.py", line 14, in _accept_method
  File "C:\Program Files\PyScripter\Lib\rpyc.zip\rpyc\utils\server.py", line 191, in _serve_client
  File "C:\Program Files\PyScripter\Lib\rpyc.zip\rpyc\core\protocol.py", line 391, in serve_all
  File "C:\Program Files\PyScripter\Lib\rpyc.zip\rpyc\core\protocol.py", line 382, in serve
  File "C:\Program Files\PyScripter\Lib\rpyc.zip\rpyc\core\protocol.py", line 350, in _dispatch
  File "C:\Program Files\PyScripter\Lib\rpyc.zip\rpyc\core\protocol.py", line 298, in _dispatch_request
  File "C:\Program Files\PyScripter\Lib\rpyc.zip\rpyc\core\protocol.py", line 528, in _handle_call
  File "<string>", line 420, in run_nodebug
  File "C:\DATA\jff\data\python\Whosebug\simple_traceback.py", line 10, in <module>
  File "C:\DATA\jff\data\python\Whosebug\simple_traceback.py", line 8, in foo
  File "C:\DATA\jff\data\python\Whosebug\simple_traceback.py", line 4, in demo

Answer 2

呸，我只是想通过允许调用者传递他们的 __file__ 值来解决这个问题:-(

我的函数：

def do_something(app, filename, relroot=None):
   if relroot is None:
      relroot = '.'
   else:
      relroot = os.path.dirname(relroot)
   path = os.path.join(relroot, filename)
   ...

在conf.py中：

def setup(app):
   mymodule.do_something(app, 'path/to/file', relroot=__file__)

有没有一种可靠的方法可以从在 Sphinx conf.py 中执行的 Python 函数中获取调用者模块的路径？

Is there a reliable way to get the path of the caller module from a Python function that is executed within a Sphinx conf.py?

python

python-sphinx