Download file from web in Python 3


I am creating a program that will download a .jar (java) file from a web server, by reading the URL that is specified in the .jad file of the same game/application. I’m using Python 3.2.1

I’ve managed to extract the URL of the JAR file from the JAD file (every JAD file contains the URL to the JAR file), but as you may imagine, the extracted value is type() string.

Here’s the relevant function:

def downloadFile(URL=None):
    import httplib2
    h = httplib2.Http(".cache")
    resp, content = h.request(URL, "GET")
    return content


However I always get an error saying that the type in the function above has to be bytes, and not string. I’ve tried using the URL.encode(‘utf-8′), and also bytes(URL,encoding=’utf-8’), but I’d always get the same or similar error.

So basically my question is how to download a file from a server when the URL is stored in a string type?

Asked By: Bo Milanovich



If you want to obtain the contents of a web page into a variable, just read the response of urllib.request.urlopen:

import urllib.request
url = ''
response = urllib.request.urlopen(url)
data =      # a `bytes` object
text = data.decode('utf-8') # a `str`; this step can't be used if data is binary

The easiest way to download and save a file is to use the urllib.request.urlretrieve function:

import urllib.request
# Download the file from `url` and save it locally under `file_name`:
urllib.request.urlretrieve(url, file_name)
import urllib.request
# Download the file from `url`, save it in a temporary directory and get the
# path to it (e.g. '/tmp/tmpb48zma.txt') in the `file_name` variable:
file_name, headers = urllib.request.urlretrieve(url)

But keep in mind that urlretrieve is considered legacy and might become deprecated (not sure why, though).

So the most correct way to do this would be to use the urllib.request.urlopen function to return a file-like object that represents an HTTP response and copy it to a real file using shutil.copyfileobj.

import urllib.request
import shutil
# Download the file from `url` and save it locally under `file_name`:
with urllib.request.urlopen(url) as response, open(file_name, 'wb') as out_file:
    shutil.copyfileobj(response, out_file)

If this seems too complicated, you may want to go simpler and store the whole download in a bytes object and then write it to a file. But this works well only for small files.

import urllib.request
# Download the file from `url` and save it locally under `file_name`:
with urllib.request.urlopen(url) as response, open(file_name, 'wb') as out_file:
    data = # a `bytes` object

It is possible to extract .gz (and maybe other formats) compressed data on the fly, but such an operation probably requires the HTTP server to support random access to the file.

import urllib.request
import gzip
# Read the first 64 bytes of the file inside the .gz archive located at `url`
url = ''
with urllib.request.urlopen(url) as response:
    with gzip.GzipFile(fileobj=response) as uncompressed:
        file_header = # a `bytes` object
        # Or do anything shown above using `uncompressed` instead of `response`.
Answered By: Oleh Prypin

I hope I understood the question right, which is: how to download a file from a server when the URL is stored in a string type?

I download files and save it locally using the below code:

import requests

url = ''
fileName = 'D:PythondwnldPythonLogo.png'
req = requests.get(url)
file = open(fileName, 'wb')
for chunk in req.iter_content(100000):
Answered By: Ranvijay Kumar

I use requests package whenever I want something related to HTTP requests because its API is very easy to start with:

first, install requests

$ pip install requests

then the code:

from requests import get  # to make GET request

def download(url, file_name):
    # open in binary mode
    with open(file_name, "wb") as file:
        # get request
        response = get(url)
        # write to file
Answered By: Ali Faki
from urllib import request

def get(url):
    with request.urlopen(url) as r:

def download(url, file=None):
    if not file:
        file = url.split('/')[-1]
    with open(file, 'wb') as f:
Answered By: user7726287

You can use wget which is popular downloading shell tool for that.
This will be the simplest method since it does not need to open up the destination file. Here is an example.

import wget
url = '', '/Users/scott/Downloads/cat4.jpg') 
Answered By: Lasith Niroshan

Here we can use urllib’s Legacy interface in Python3:

The following functions and classes are ported from the Python 2 module urllib (as opposed to urllib2). They might become deprecated at some point in the future.

Example (2 lines code):

import urllib.request

url = ''
urllib.request.urlretrieve(url, "logo.png")
Answered By: Yang Yu

Yes, definietly requests is great package to use in something related to HTTP requests. but we need to be careful with the encoding type of the incoming data as well below is an example which explains the difference

from requests import get

# case when the response is byte array
url = 'some_image_url'

response = get(url)
with open('output', 'wb') as file:

# case when the response is text
# Here unlikely if the reponse content is of type **iso-8859-1** we will have to override the response encoding
url = 'some_page_url'

response = get(url)
# override encoding by real educated guess as provided by chardet
r.encoding = r.apparent_encoding

with open('output', 'w', encoding='utf-8') as file:

Answered By: Development Lead


Sometimes, we are want to get the picture but not need to download it to real files,

i.e., download the data and keep it on memory.

For example, If I use the machine learning method, train a model that can recognize an image with the number (bar code).

When I spider some websites and that have those images so I can use the model to recognize it,

and I don’t want to save those pictures on my disk drive,

then you can try the below method to help you keep download data on memory.


import requests
from io import BytesIO
response = requests.get(url)
with BytesIO as io_obj:
    for chunk in response.iter_content(chunk_size=4096):

basically, is like to @Ranvijay Kumar

An Example

import requests
from typing import NewType, TypeVar
from io import StringIO, BytesIO
import matplotlib.pyplot as plt
import imageio

URL = NewType('URL', str)
T_IO = TypeVar('T_IO', StringIO, BytesIO)

def download_and_keep_on_memory(url: URL, headers=None, timeout=None, **option) -> T_IO:
    chunk_size = option.get('chunk_size', 4096)  # default 4KB
    max_size = 1024 ** 2 * option.get('max_size', -1)  # MB, default will ignore.
    response = requests.get(url, headers=headers, timeout=timeout)
    if response.status_code != 200:
        raise requests.ConnectionError(f'{response.status_code}')

    instance_io = StringIO if isinstance(next(response.iter_content(chunk_size=1)), str) else BytesIO
    io_obj = instance_io()
    cur_size = 0
    for chunk in response.iter_content(chunk_size=chunk_size):
        cur_size += chunk_size
        if 0 < max_size < cur_size:
    """ save it to real file.
    with open('temp.png', mode='wb') as out_f:
    return io_obj

def main():
    headers = {
        'Accept': 'text/html,application/xhtml+xml,application/xml;q=0.9,image/webp,image/apng,*/*;q=0.8,application/signed-exchange;v=b3',
        'Accept-Encoding': 'gzip, deflate',
        'Accept-Language': 'zh-TW,zh;q=0.9,en-US;q=0.8,en;q=0.7',
        'Cache-Control': 'max-age=0',
        'Connection': 'keep-alive',
        'Host': '',
        'Upgrade-Insecure-Requests': '1',
        'User-Agent': 'Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/78.0.3904.87 Safari/537.36'
    io_img = download_and_keep_on_memory(URL(''),
                                         headers,  # You may need this. Otherwise, some websites will send the 404 error to you.
                                         max_size=4)  # max loading < 4MB
    with io_img:
        plt.rc('axes.spines', top=False, bottom=False, left=False, right=False)
        plt.rc(('xtick', 'ytick'), color=(1, 1, 1, 0))  # same of plt.axis('off')
        plt.imshow(imageio.imread(io_img, as_gray=False, pilmode="RGB"))

if __name__ == '__main__':

Answered By: Carson

If you are using Linux you can use the wget module of Linux through the python shell. Here is a sample code snippet

import os
url = ''
os.system('wget %s'%url)
Answered By: Gaurav Shrivastava
Categories: questions Tags: ,
Answers are sorted by their score. The answer accepted by the question owner as the best is marked with
at the top-right corner.