How can I read the contents of an URL with Python?

To answer your question:

import urllib

link = "http://www.somesite.com/details.pl?urn=2344"
f = urllib.urlopen(link)
myfile = f.read()
print(myfile)

You need to read(), not readline()

EDIT (2018-06-25): Since Python 3, the legacy urllib.urlopen() was replaced by urllib.request.urlopen() (see notes from https://docs.python.org/3/library/urllib.request.html#urllib.request.urlopen for details).

If you're using Python 3, see answers by Martin Thoma or i.n.n.m within this question: https://stackoverflow.com/a/28040508/158111 (Python 2/3 compat) https://stackoverflow.com/a/45886824/158111 (Python 3)

Or, just get this library here: http://docs.python-requests.org/en/latest/ and seriously use it :)

import requests

link = "http://www.somesite.com/details.pl?urn=2344"
f = requests.get(link)
print(f.text)

A solution with works with Python 2.X and Python 3.X makes use of the Python 2 and 3 compatibility library six:

from six.moves.urllib.request import urlopen
link = "http://www.somesite.com/details.pl?urn=2344"
response = urlopen(link)
content = response.read()
print(content)

For python3 users, to save time, use the following code,

from urllib.request import urlopen

link = "https://docs.scipy.org/doc/numpy/user/basics.broadcasting.html"

f = urlopen(link)
myfile = f.read()
print(myfile)

I know there are different threads for error: Name Error: urlopen is not defined, but thought this might save time.

None of these answers are very good for Python 3 (tested on latest version at the time of this post).

This is how you do it...

import urllib.request

try:
   with urllib.request.urlopen('http://www.python.org/') as f:
      print(f.read().decode('utf-8'))
except urllib.error.URLError as e:
   print(e.reason)

The above is for contents that return 'utf-8'. Remove .decode('utf-8') if you want python to "guess the appropriate encoding."

Documentation: https://docs.python.org/3/library/urllib.request.html#module-urllib.request