sending signals to the calling function

Tim Golden tim.golden at
Mon Jun 28 11:24:31 CEST 2004

| I have a function that receive a list of files and creates a zip from
| these files. I want it to send signal to the calling function with the
| name of the file it currently compressing. is there a way to do this
| (without threads)?

Of course, just pass in a callback function, and call that
whenever a file is being compressed. Alternatively, make
the zipper-upper a generator which yields its current filename:


import os, sys
import zipfile

def zipup (zip_filename, list_of_files):
    archive = zipfile.ZipFile (zip_filename, "w")
    for filename in list_of_files:
        yield filename
        archive.write (filename)
    archive.close ()

if __name__ == '__main__':
    import glob
    file_pattern = os.path.join (sys.path[-1], "*")
    list_of_files = [f for f in glob.glob (file_pattern) if os.path.isfile
    for zipping_file in zipup ("", list_of_files):
        print "About to zip", zipping_file



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