Python - Open a file with a wildcard (%) directory path in Windows

Question

In Python, I'm trying to open a file that gets saved to the %TEMP% directory. I've tried:

file = open("%TEMP%\file.txt")

and

file = open("%%TEMP%%\file.txt")

and

file = open("%TEMP%\\file.txt")

and

file = open("%%TEMP%%\\file.txt")

And always get (this one specifically for that last example):

IOError: [Errno 2] No such file or directory: '%%TEMP%%\\file.txt'

For sanity's sake, from Windows command prompt I do a type %TEMP%\\file.txt and it prints out the file OK. Any help?

Answer 1

Use os.environ

import os
f = open(os.path.join(os.environ['TEMP'], 'file.txt'))

You can also use os.path.expandvars

import os
f = open(os.path.expandvars(r'%TEMP%\file.txt'))