sending signals to the calling function

[Haim Ashkenazi]

On Mon, 2004-06-28 at 12:24, Tim Golden wrote:
> | 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:
> 
> <code>
> 
> 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
> (f)]
>     for zipping_file in zipup ("test.zip", list_of_files):
>         print "About to zip", zipping_file
thanx, that's exactly what I needed.

Bye
-- 
Haim
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