> /main/parallel_branch_1/release_branch_1.0/dbg_for_python/CHECKEDOUT
> from /main/parallel_branch_1/release_branch_1.0/4
>
> I want to write a regex that gives me the branch the file was
> checkedout on ,in this case - 'dbg_for_python'
>
> Also if there is a better way than using regex, please let me know.
Well, if you have it all in a single string:
s =
"/main/parallel_branch_1/release_branch_1.0/dbg_for_python/CHECKEDOUT
from /main/parallel_branch_1/release_branch_1.0/4"
you can do
branch = s.split("/")[4]
which returns the branch, assuming the path from root is the
same for each item in question.
If not, you can tinker with something like
r = re.compile(r'/([^/]*)/CHECKEDOUT')
m = r.match(s)
and which should make m.groups(1) the resulting item. You
don't give much detail regarding what is constant (the
number of subdirectories in the path? the CHECKEDOUT
portion?, etc) so it's kinda hard to figure out what is most
globally applicable.
-tkc
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