31 lines
1.1 KiB
Python
31 lines
1.1 KiB
Python
|
'''
|
||
|
|
This utility allows a python system to find a file in it's
|
||
|
|
directory. To do this, you need to pass it a function object from
|
||
|
|
a module in the correct directory. I know there must be a better
|
||
|
|
way to do this, but I haven't seen it yet. Incidentally, the
|
||
|
|
current directory should be _different_ from the module in which
|
||
|
|
the function is contained, otherwise this function will go off into
|
||
|
|
the root directory.
|
||
|
|
|
||
|
|
Currently this has to be called with the current directory a directory
|
||
|
|
other than the directory we're trying to find... need a better solution
|
||
|
|
for this kind of thing... a python registry would be great :)
|
||
|
|
|
||
|
|
NOTE: as of Python 1.5, this module should be obsolete! As soon as I
|
||
|
|
have verified that all of my code is fixed, it will be moved to the unused
|
||
|
|
directories.
|
||
|
|
'''
|
||
|
|
import os,sys
|
||
|
|
|
||
|
|
def findourfile(function, filename):
|
||
|
|
'''
|
||
|
|
Given the function, return a path to the a file in the
|
||
|
|
same directory with 'filename'. We also let the caller
|
||
|
|
know if the file already exists.
|
||
|
|
'''
|
||
|
|
ourfilename = os.path.split(function.func_code.co_filename)[0]+os.sep+filename
|
||
|
|
exists = os.path.exists(ourfilename)
|
||
|
|
return (exists,ourfilename)
|
||
|
|
|
||
|
|
|