Simple UnZip Script, help

Hi All,

I am looking for help on this simple script to unzip/extract the contents of a zip file. This is what I have so far:

Code: ( text )

import zipfile, os, sys
zip1 = ("C:\\Temp\\test11.zip")
z = zipfile.ZipFile(zip1, '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!"

I had the unzipping working, but not anymore. As is, the script executes w/o error, but doesn't unzip the files. Also, how would you unzip the file to a different directory? Is there a better way to do this then what I'm doing? I got much of this code from a post elsewhere, but am not too sure I follow it.

Thanks!

kaarthikeyapreyan

Member

72 Posts

July 2nd, 2008
10:59 AM
#2

Re: Simple UnZip Script, help

Try this code, hope this suits your requirement.

Code: ( text )

import os, sys, zipfile
class myzip:
def __init__(self):
source = raw_input('Enter the name of the zipfile you want to extract :')
self.diffdir = raw_input('Enter the name of the dir you want to extract the file if in present dir enter "." :')
fh = open(source, 'rb')
self.z = zipfile.ZipFile(fh)
dirs = []
for name in self.z.namelist():
if name.endswith('/'):
dirs.append(name)
self._checkdiff(dirs)
def _checkdiff(self,dirs):
"""
Appends if different directory is specified
"""
for dirname in dirs:
if self.diffdir == "." :
self.flag = 0
else :
dirname = os.path.join(self.diffdir,dirname)
self.flag = 1
self.generatedir(dirname)
self.generatefiles()
def generatedir(self,dirname):
"""
Generate the directories for our new zip archive
"""
if not os.path.isdir(dirname):
os.mkdir(dirname)
else:
print "Directory exist"
sys.exit(1)
def generatefiles(self):
"""
Write files from archive to existence
"""
for name in self.z.namelist():
if self.flag == 1:
dest = os.path.join(self.diffdir,name)
if not dest.endswith('/'):
outfile = open(dest, 'wb')
outfile.write(self.z.read(name))
outfile.close()
self.z.close()
obj = myzip()

jld730

Member

33 Posts

July 2nd, 2008
02:22 PM
#3

Re: Simple UnZip Script, help

Thanks, that is helpful, especially for future use. For my specifc/current needs, I went with this:

Code: ( text )

import zipfile, os, sys
test1_zip = ("C:\\Temp\\test1.zip")
folder_name = sys.argv[1]
output_dir = "C:\\Temp\\" + folder_name
z = zipfile.ZipFile(test1_zip, 'r')
zList = z.namelist()
for zItem in zList:
print "Unpacking",zItem
zRead = z.read(zItem)
z1File = open(os.path.join(output_dir, zItem),'wb')
z1File.write(zRead)
z1File.close
print "Finished"

heiro

Member

51 Posts

July 3rd, 2008
05:57 PM
#4

Re: Simple UnZip Script, help

Hi,

if you want to preserve the time stamp of the file you can also use this code.
This can also unpack zip files with folders.

Code: ( text )

def unzip(src,dst,name=''):
if not os.path.isfile(src):
print 'Zip file not found'
else:
archive = ZipFile(src,'r')
#src = os.path.basename(src)
if name=='':
name = os.path.basename(os.path.dirname(dst))
dst=dst+name+'/'
for file in archive.namelist():
try:
os.makedirs(normpath((abspath(dst)+'/'+dirname(file))))
except:
pass
if not str(normpath((abspath(dst)+'/'+dirname(file)))+"/"+basename(file)).endswith('/'):
efile = open(normpath((abspath(dst)+'/'+dirname(file)))+"/"+basename(file),'wb')
efile.write(archive.read(file))
efile.close()
accesstime = time.time()
timeTuple=(int(archive.getinfo(file).date_time[0]),\
int(archive.getinfo(file).date_time[1]),\
int(archive.getinfo(file).date_time[2]),\
int(archive.getinfo(file).date_time[3]) ,\
int(archive.getinfo(file).date_time[4]),\
int(archive.getinfo(file).date_time[5]),\
int(0),int(0),int(0))
modifiedtime = mktime(timeTuple)
try:
utime((dst+file), (accesstime,modifiedtime))
except:
pass
archive.close()