[Tutor] copy zip file from source folder to destination and unzip all files within destination

[Kent Johnson]

On Fri, Nov 20, 2009 at 1:54 AM, MARCUS NG <markersng at gmail.com> wrote:
> Hey all,
> I have been searching online for ways to copy a zip file to a destination
> and extract the zip file with python.
> Currently nothing works due to my limited understanding.
> I am wondering if my syntax is wrong or am I missing anything?

You don't say how it fails but you should at least escape the \ in the
paths as \\ or use raw strings (r'C:\Users...')

The \char codes are escape codes so your paths are not correct.

Kent

> the code is as such. also if there is a better code, I am definitely open to
> it. If it is good, can you also explain steps?
> ###########
>
> import shutil, zipfile
> shutil.copyfile('C:\Users\blueman\Desktop\myTest1.0.zip',
> 'C:\Users\blueman\Desktop\allFiles')
>
> zipfilepath='C:\Users\blueman\Desktop\allFiles\myTest1.0.zip'
> extractiondir='C:\Users\blueman\Desktop\allFiles\test'
>
> def extract(zipfilepath, extractiondir):
>     zip = zipfile(zipfilepath)
>     zip.extractall(path=extractiondir)
>     print 'it works'
>
> extract()
>
> ###########
> thank you so much for your time and help in advance!!!
>
>
> ,
> Marcus
>
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