Download Returned Zip file from URL

Each Answer to this Q is separated by one/two green lines.

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?

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))

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))

Most people recommend using requests if it is available, and the requests documentation recommends this for downloading and saving raw data from a url:

import requests 

def download_url(url, save_path, chunk_size=128):
    r = requests.get(url, stream=True)
    with open(save_path, 'wb') as fd:
        for chunk in r.iter_content(chunk_size=chunk_size):

Since the answer asks about downloading and saving the zip file, I haven’t gone into details regarding reading the zip file. See one of the many answers below for possibilities.

If for some reason you don’t have access to requests, you can use urllib.request instead. It may not be quite as robust as the above.

import urllib.request

def download_url(url, save_path):
    with urllib.request.urlopen(url) as dl_file:
        with open(save_path, 'wb') as out_file:

Finally, if you are using Python 2 still, you can use urllib2.urlopen.

from contextlib import closing

def download_url(url, save_path):
    with closing(urllib2.urlopen(url)) as dl_file:
        with open(save_path, 'wb') as out_file:

With the help of this blog post, I’ve got it working with just requests.
The point of the weird stream thing is so we don’t need to call content
on large requests, which would require it to all be processed at once,
clogging the memory. The stream avoids this by iterating through the data
one chunk at a time.


response = requests.get(url, stream=True)
with open('', "wb") as f:
    for chunk in response.iter_content(chunk_size=512):
        if chunk:  # filter out keep-alive new chunks

Here’s what I got to work in Python 3:

import zipfile, urllib.request, shutil


with urllib.request.urlopen(url) as response, open(file_name, 'wb') as out_file:
    shutil.copyfileobj(response, out_file)
    with zipfile.ZipFile(file_name) as zf:

Either use urllib2.urlopen, or you could try using the excellent Requests module and avoid urllib2 headaches:

import requests
results = requests.get('url')
#pass results.content onto secondary processing...

I came here searching how to save a .bzip2 file. Let me paste the code for others who might come looking for this.

url = ""
filename = "swateek.tar.gz"

response = requests.get(url, headers=headers, auth=('myusername', 'mypassword'), timeout=50)
if response.status_code == 200:
with open(filename, 'wb') as f:

I just wanted to save the file as is.

Super lightweight solution to save a .zip file to a location on disk (using Python 3.9):

import requests

url = r'https://linktofile'
output = r'C:\pathtofolder\'

r = requests.get(url)
with open(output, 'wb') as f:

Thanks to @yoavram for the above solution,
my url path linked to a zipped folder, and encounter an error of BADZipfile
(file is not a zip file), and it was strange if I tried several times it
retrieve the url and unzipped it all of sudden so I amend the solution a little
bit. using the is_zipfile method as per here

r = requests.get(url, stream =True)
check = zipfile.is_zipfile(io.BytesIO(r.content))
while not check:
    r = requests.get(url, stream =True)
    check = zipfile.is_zipfile(io.BytesIO(r.content))
    z = zipfile.ZipFile(io.BytesIO(r.content))

Use requests, zipfile and io python packages.

Specially BytesIO function is used to keep the unzipped file in memory rather than saving it into the drive.

import requests
from zipfile import ZipFile
from io import BytesIO

r = requests.get(zip_file_url)
z = ZipFile(BytesIO(r.content))    
file = z.extract(a_file_to_extract, path_to_save)
with open(file) as f:

The answers/resolutions are collected from stackoverflow, are licensed under cc by-sa 2.5 , cc by-sa 3.0 and cc by-sa 4.0 .

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