Python如何输出环境变量

sunny03

　　最近项目进展顺利，编码上面没干什么事情，就学起了Python
　　在wxpython遇到一个问题，ubuntu 12.04下面MenuBar没办法显示
　　查询了伟大的stackoverflow，证明是Ubuntu的一个bug，在调用脚本前设置环境变量就可以解决

export UBUNTU_MENUPROXY=0
　　就想着集成到脚本里面去，自然的想到如下代码

import os
os.system("export UBUNTU_MENUPROXY=0")
　　但是it doesn't work! why? 还是求助 stackoverflow，看到如下内容

export is a command that you give directly to the shell (e.g. bash), to tell it to add or modify one of its environment variables. You can't change your shell's environment from a child process (such as Python), it's just not possible.
Here's what's happening with you try os.system('export MY_DATA="my_export"')...
/bin/bash process, command `python yourscript.py` forks python subprocess
|_
/usr/bin/python process, command `os.system()` forks /bin/sh subprocess
|_
/bin/sh process, command `export ...` changes local environment
When the bottom-most /bin/sh subprocess finishes running your export ... command, then it's discarded, along with the environment that you have just changed.
　　大牛们说了，这段的线程树是这样的
　　/bin/bash -> /usr/bin/python -> /bin/sh
　　export 是在/bin/sh中执行的，执行完成后就退出这个子线程了，export了解的话就知道export是export到当前线程，对父线程不可见，于是在/usr/bin/python中当然起不到作用？
　　正确的设置方法是？ 很简单：

import os
os.environ['UBUNTU_MENUPROXY'] = 0
　　搞定