If it won't be simple, it simply won't be. [Hire me, source code] by Miki Tebeka, CEO, 353Solutions

Sunday, July 20, 2008


#!/usr/bin/env python
'''Find under which SCM directory is'''

__author__ = "Miki Tebeka <miki.tebeka@gmail.com>"

from os import sep
from os.path import join, isdir, abspath
from itertools import ifilter, imap

def updirs(path):
parts = path.split(sep)
if not parts[0]:
parts[0] = sep # FIXME: Windows
while parts:
yield join(*parts)

def scmdirs(path, scms):
for scmext in scms:
yield join(path, scmext)

def scm(dirname):
return dirname[-3:].lower()

def scms(path, scms):
return imap(scm, ifilter(isdir, scmdirs(path, scms)))

def whichscm(path):
path = abspath(path)

for scm in scms(path, (".svn", "CVS")):
return scm

scmdirs = (".bzr", ".hg", ".git")
for dirname in updirs(path):
for scm in scms(dirname, (".bzr", ".hg", ".git")):
return scm

def main(argv=None):
if argv is None:
import sys
argv = sys.argv

from optparse import OptionParser

parser = OptionParser("usage: %prog [DIRNAME]")

opts, args = parser.parse_args(argv[1:])
if len(args) not in (0, 1):
parser.error("wrong number of arguments") # Will exit

dirname = args[0] if args else "."
if not isdir(dirname):
raise SystemExit("error: %s is not a directory" % dirname)

scm = whichscm(dirname)
if not scm:
raise SystemExit("error: can't find scm for %s" % dirname)

print scm

if __name__ == "__main__":

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