How do I download a file over HTTP using Python?


Question

I have a small utility that I use to download a MP3 from a website on a schedule and then builds/updates a podcast XML file which I've obviously added to iTunes.

The text processing that creates/updates the XML file is written in Python. I use wget inside a Windows .bat file to download the actual MP3 however. I would prefer to have the entire utility written in Python though.

I struggled though to find a way to actually down load the file in Python, thus why I resorted to wget.

So, how do I download the file using Python?

1
797
3/31/2017 5:11:20 AM

Accepted Answer

In Python 2, use urllib2 which comes with the standard library.

import urllib2
response = urllib2.urlopen('http://www.example.com/')
html = response.read()

This is the most basic way to use the library, minus any error handling. You can also do more complex stuff such as changing headers. The documentation can be found here.

433
6/5/2014 11:06:21 AM

One more, using urlretrieve:

import urllib
urllib.urlretrieve ("http://www.example.com/songs/mp3.mp3", "mp3.mp3")

(for Python 3+ use import urllib.request and urllib.request.urlretrieve)

Yet another one, with a "progressbar"

import urllib2

url = "http://download.thinkbroadband.com/10MB.zip"

file_name = url.split('/')[-1]
u = urllib2.urlopen(url)
f = open(file_name, 'wb')
meta = u.info()
file_size = int(meta.getheaders("Content-Length")[0])
print "Downloading: %s Bytes: %s" % (file_name, file_size)

file_size_dl = 0
block_sz = 8192
while True:
    buffer = u.read(block_sz)
    if not buffer:
        break

    file_size_dl += len(buffer)
    f.write(buffer)
    status = r"%10d  [%3.2f%%]" % (file_size_dl, file_size_dl * 100. / file_size)
    status = status + chr(8)*(len(status)+1)
    print status,

f.close()

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