download a zip file to a local drive and extract all files to a destination folder using python 2.5

问题

I am trying to download a zip file to a local drive and extract all files to a destination folder.

so i have come up with solution but it is only to "download" a file from a directory to another directory but it doesn't work for downloading files. for the extraction, I am able to get it to work in 2.6 but not for 2.5. so any suggestions for the work around or another approach I am definitely open to. thanks in advance.

######################################
'''this part works but it is not good for URl links''' 
import shutil

sourceFile = r"C:\Users\blueman\master\test2.5.zip"
destDir = r"C:\Users\blueman\user"
shutil.copy(sourceFile, destDir)
print "file copied"
######################################################

'''extract works but not good for version 2.5'''
import zipfile

GLBzipFilePath =r'C:\Users\blueman\user\test2.5.zip'
GLBextractDir =r'C:\Users\blueman\user'

def extract(zipFilePath, extractDir):
 zip = zipfile(zipFilePath)
 zip.extractall(path=extractDir)
 print "it works"

extract(GLBzipFilePath,GLBextractDir)

######################################################

回答1:

urllib.urlretrieve can get a file (zip or otherwise;-) from a URL to a given path.

extractall is indeed new in 2.6, but in 2.5 you can use an explicit loop (get all names, open each name, etc). Do you need example code?

So here's the general idea (needs more try/except if you want to give a nice error message in each and every case which could go wrong, of which, of course, there are a million variants -- I'm only using a couple of such cases as examples...):

import os
import urllib
import zipfile

def getunzipped(theurl, thedir):
  name = os.path.join(thedir, 'temp.zip')
  try:
    name, hdrs = urllib.urlretrieve(theurl, name)
  except IOError, e:
    print "Can't retrieve %r to %r: %s" % (theurl, thedir, e)
    return
  try:
    z = zipfile.ZipFile(name)
  except zipfile.error, e:
    print "Bad zipfile (from %r): %s" % (theurl, e)
    return
  for n in z.namelist():
    dest = os.path.join(thedir, n)
    destdir = os.path.dirname(dest)
    if not os.path.isdir(destdir):
      os.makedirs(destdir)
    data = z.read(n)
    f = open(dest, 'w')
    f.write(data)
    f.close()
  z.close()
  os.unlink(name)

回答2:

For downloading, look at urllib:

import urllib
webFile = urllib.urlopen(url)

For unzipping, use zipfile. See also this example.

回答3:

The shortest way i've found so far, is to use +alex answer, but with ZipFile.extractall() instead of the loop:

from zipfile import ZipFile
from urllib import urlretrieve
from tempfile import mktemp

filename = mktemp('.zip')
destDir = mktemp()
theurl = 'http://www.example.com/file.zip'
name, hdrs = urlretrieve(theurl, filename)
thefile=ZipFile(filename)
thefile.extractall(destDir)
thefile.close()

来源：https://stackoverflow.com/questions/1774434/download-a-zip-file-to-a-local-drive-and-extract-all-files-to-a-destination-fold

标签

python

http

download

extract

unzip