Download Returned Zip file from URL

python url download zip urllib
user1229108 picture user1229108 · Feb 23, 2012 · Viewed 115.8k times · Source

If I have a URL that, when submitted in a web browser, pops up a dialog box to save a zip file, how would I go about catching and downloading this zip file in Python?

Answer

yoavram picture yoavram · Jan 10, 2013

As far as I can tell, the proper way to do this is:

import requests, zipfile, StringIO
r = requests.get(zip_file_url, stream=True)
z = zipfile.ZipFile(StringIO.StringIO(r.content))
z.extractall()

of course you'd want to check that the GET was successful with r.ok.

For python 3+, sub the StringIO module with the io module and use BytesIO instead of StringIO: Here are release notes that mention this change.

import requests, zipfile, io
r = requests.get(zip_file_url)
z = zipfile.ZipFile(io.BytesIO(r.content))
z.extractall("/path/to/destination_directory")

Related questions

How do I send a POST request as a JSON?

data = { 'ids': [12, 3, 4, 5, 6 , ...] } urllib2.urlopen("http://abc.com/api/posts/create",urllib.urlencode(data)) I want to send a POST request, but one of the fields should be a list of numbers. How can I do that ? (JSON?)

Read More
How to percent-encode URL parameters in Python?

If I do url = "http://example.com?p=" + urllib.quote(query) It doesn't encode / to %2F (breaks OAuth normalization) It doesn't handle Unicode (it throws an exception) Is there a better library?

Read More
How to validate a url in Python? (Malformed or not)

I have url from the user and I have to reply with the fetched HTML. How can I check for the URL to be malformed or not? For example : url='google' // Malformed url='google.com' // Malformed url='http://google.com' // …

Read More