os.path pathname manipulations

[Bouzy]

for dirpath,dirs,files in os.walk(cwd):
for file in files:

global z_path
z_path = os.path.  # WHAT WOULD GO HERE #  (dirpath)   

print z_path

This is part of a script where I am tying to grab the names of folders and an undetermined number of sub-folders/files and print them to screen. I don't know how deep the sub-folders will go. Currently when I do this it prints things like

C:\Documents and Settings\B\Desk top\Test\T1\tes t.txt.

What I want is to go from cwd (in this case Test) and just print to screen

\T1\test.txt

I have tried many os.path pathname manipulators like...

abspath( path)
basename( path)
commonprefix( list)
dirname( path)
split( path)
splitdrive( path)
splitunc( path)

...and none seem to work.

I can't think of any way to do this with a url that will change so the url could also be...

C:\Documents and Settings\B\Desk top\Test\T1\444 \test.txt.

and I want it to go from test > test.txt or T1\444\test.txt ...

Ignore whats below if it doesn't make sense.
I thought of assigning a variable to cwd and then splitting the url based on that, but I don't know how to go about doing that specifically.

Does anyone know how this could be done?

[nishi2rock]

I could not exactly understand your problem.
According to my understanding you dont want the whole Filepath to be printed on the console.

Try this code

Curdir = os.getcwd()

for DirInfo in os.walk(Curdir):
    for File in DirInfo[2]:
       print (DirInfo[0]+"\\"+File).replace(Curdir,"")