Download file from web in Python 3

Question

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 usi

Answer 1

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
        file.write(response.content)

Answer 2

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:
    file.write(response.content)


# 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:
    file.write(response.content)

Answer 3

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 = 'http://example.com/'
response = urllib.request.urlopen(url)
data = response.read()      # 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 = response.read() # a `bytes` object
    out_file.write(data)

---

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 = 'http://example.com/something.gz'
with urllib.request.urlopen(url) as response:
    with gzip.GzipFile(fileobj=response) as uncompressed:
        file_header = uncompressed.read(64) # a `bytes` object
        # Or do anything shown above using `uncompressed` instead of `response`.