如何使用Python通过HTTP读取远程Zip存档中的选定文件？

Question

I need to read selected files, matching on the file name, from a remote zip archive using Python. I don't want to save the full zip to a temporary file (it's not that large, so I can handle everything in memory).

I've already written the code and it works, and I'm answering this myself so I can search for it later. But since evidence suggests that I'm one of the dumber participants on Stackoverflow, I'm sure there's room for improvement.

Answer 1

Here's how I did it (grabbing all files ending in ".ranks"):

import urllib2, cStringIO, zipfile

try:
    remotezip = urllib2.urlopen(url)
    zipinmemory = cStringIO.StringIO(remotezip.read())
    zip = zipfile.ZipFile(zipinmemory)
    for fn in zip.namelist():
        if fn.endswith(".ranks"):
            ranks_data = zip.read(fn)
            for line in ranks_data.split("\n"):
                # do something with each line
except urllib2.HTTPError:
    # handle exception

Answer 2

This will do the job without downloading the entire zip file!

http://pypi.python.org/pypi/pyremotezip

Answer 3

Thanks Marcel for your question and answer (I had the same problem in a different context and encountered the same difficulty with file-like objects not really being file-like)! Just as an update: For Python 3.0, your code needs to be modified slightly:

import urllib.request, io, zipfile

try:
    remotezip = urllib.request.urlopen(url)
    zipinmemory = io.BytesIO(remotezip.read())
    zip = zipfile.ZipFile(zipinmemory)
    for fn in zip.namelist():
        if fn.endswith(".ranks"):
            ranks_data = zip.read(fn)
            for line in ranks_data.split("\n"):
                # do something with each line
except urllib.request.HTTPError:
    # handle exception

Answer 4

请记住，仅解压缩ZIP文件可能会导致安全漏洞 。