simple UnZip

[Cédric Lucantis]

Le Wednesday 02 July 2008 15:39:51 noydb, vous avez écrit :
> Can someone help me with this script, which I found posted elsewhere?
> I'm trying to figure out what is going on here so that I can alter it
> for my needs, but the lack of descriptive names is making it
> difficult.  And, the script doesn't quite do anything worthwhile -- it
> unzips one file from a zipfile, not all files in a zipfile.
>
> ***
> import zipfile, os, sys, glob
>
> os.chdir("C:\\Temp")
> zips = glob.glob('*.zip')
>
> for fzip in zips:
>     if zipfile.is_zipfile(fzip):
>         print fzip," is a zip"
>         z = zipfile.ZipFile(fzip,'r')
>         lstName = z.namelist()
>         sHgt = lstName[0]
>         print "Unpacking",sHgt
>         hgt = z.read(sHgt)
>         fHgt = open(sHgt,'wb')
>         fHgt.write(hgt)
>         # fHgt.flush
>         fHgt.close
> print "Finished"
> ***
>
> I changed it somewhat to
> &&&
> import zipfile, os, sys
>
> event_zip = ("C:\\Temp\\data4event.zip")
>
> z = zipfile.ZipFile(event_zip, 'r')
>
> zList = z.namelist()
>
> for zItem in zList:
>     print "Unpacking",zItem
>     zRead = z.read(zItem)
>     z1File = open(zItem,'wb')
>     z1File.write(zRead)
>     z1File.close

namelist() returns a list of relative file names, so you can just put them 
anywhere you want with:

zlFile = open(os.path.join(DESTDIR, zItem), 'wb')

or change the current directory, but the first way should be preferred.

> print "Finished"
> &&&
>
> This works, but I want to be able to specify a different output
> location.
>

-- 
Cédric Lucantis