simple UnZip

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
print "Finished"
&&&

This works, but I want to be able to specify a different output
location.

The scenario is that the zip file will always be the same (gets copied
over daily), but it needs to be unzipped to a specific different
directory.

Can anyone help?

Thanks!

re: simple UnZip

On Jul 2, 2:39*pm, noydb <jenn.du...@gmail.comwrote:Firstly, I'd recommend just reading the documentation for the zipfile
module, it's fairly straight forwards.

To write the files to a different location, just change this line:

z1File = open(zItem,'wb')

This is where you're opening a file to write to.

Try something like:

z1File = open( os.path.join(output_dir, zItem), 'wb')

to write the files into the path specified in output_dir.

You don't even have to save the files with the same names they have in
the zipfile; you don't have to save the files at all. In one project
I'm working on, I just read files from a zip file into memory and
process them there.

re: simple UnZip

On Jul 2, 3:39*pm, noydb <jenn.du...@gmail.comwrote:Well, how is the output directory different ? Is it just a timestamp
like 'YYYYMMDD' that obviously changes every day or is it named after
the file used ? If it is timestamp style you can do

import zipfile, os, sys, time

folder_name = time.strftime("%Y%m%d", time.localtime(time.time()))
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(os.path.join(folder_name, zItem),'wb')
z1File.write(zRead)
z1File.close
print "Finished"

That will create a folder in your current working directory with the
name of todays date and unzip it there. This is just one example
route.

re: simple UnZip

Le Wednesday 02 July 2008 15:39:51 noydb, vous avez écrit*: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.
--
Cédric Lucantis

re: simple UnZip

On Jul 2, 10:07*am, Cédric Lucantis <o...@no-log.orgwrote:Thanks everyone! I did read the help on zipfile. I know it was
simple, but for newbies, sometimes it just helps to see how it is done
and learn from that.

re: simple UnZip

On Jul 2, 3:07*pm, Cédric Lucantis <o...@no-log.orgwrote:It's not actually closing the file. It should be:

z1File.close()