是否有不解引用符号链接的os.getcwd()版本? [英] Is there a version of os.getcwd() that doesn't dereference symlinks?

问题描述

可能重复:
如何在python中获取/设置逻辑目录路径

Possible Duplicate:
How to get/set logical directory path in python

我有一个从符号链接目录运行的Python脚本，我在其中调用os.getcwd()，以期获得我运行它的符号链接路径.相反，它为我提供了真实"的路径，在这种情况下这没有帮助.我需要它才能真正给我符号链接的版本.

I have a Python script that I run from a symlinked directory, and I call os.getcwd() in it, expecting to get the symlinked path I ran it from. Instead it gives me the "real" path, and in this case that's not helpful. I need it to actually give me the symlinked version.

Python对此有命令吗?

Does Python have a command for that?

推荐答案

解决方法:os.getenv('PWD')

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