How to handle urllib's timeout in Python 3?

The exception is timeout from socket, so

from socket import timeout
try:
    response = urllib.request.urlopen(url, timeout=10).read().decode('utf-8')
except (HTTPError, URLError) as error:
    logging.error('Data of %s not retrieved because %s\nURL: %s', name, error, url)
except timeout:
    logging.error('socket timed out - URL %s', url)
else:
    logging.info('Access successful.')

should catch the new exception.

The previous answer does not correctly intercept timeout errors. Timeout errors are raised as URLError, so if we want to specifically catch them, we need to write:

from urllib.error import HTTPError, URLError
import socket

try:
    response = urllib.request.urlopen(url, timeout=10).read().decode('utf-8')
except HTTPError as error:
    logging.error('Data not retrieved because %s\nURL: %s', error, url)
except URLError as error:
    if isinstance(error.reason, socket.timeout):
        logging.error('socket timed out - URL %s', url)
    else:
        logging.error('some other error happened)
else:
    logging.info('Access successful.')

Note that ValueError can independently be raised, i.e. if the URL is invalid. Like HTTPError, it is not associated with a timeout.