How to retry urllib2.request when fails?

问题

When urllib2.request reaches timeout, a urllib2.URLError exception is raised. What is the pythonic way to retry establishing a connection?

回答1:

I would use a retry decorator. There are other ones out there, but this one works pretty well. Here's how you can use it:

@retry(urllib2.URLError, tries=4, delay=3, backoff=2)
def urlopen_with_retry():
    return urllib2.urlopen("http://example.com")

This will retry the function if URLError is raised. Check the link above for documentation on the parameters, but basically it will retry a maximum of 4 times, with an exponential backoff delay doubling each time, e.g. 3 seconds, 6 seconds, 12 seconds.

回答2:

There are a few libraries out there that specialize in this.

One is backoff, which is designed with a particularly functional sensibility. Decorators are passed arbitrary callables returning generators which yield successive delay values. A simple exponential backoff with a maximum retry time of 32 seconds could be defined as:

@backoff.on_exception(backoff.expo,
                      urllib2.URLError,
                      max_value=32)
def url_open(url):
    return urllib2.urlopen("http://example.com")

Another is retrying which has very similar functionality but an API where retry parameters are specified by way of predefined keyword args.

回答3:

To retry on timeout you could catch the exception as @Karl Barker suggested in the comment:

assert ntries >= 1
for _ in range(ntries):
    try:
        page = urlopen(request, timeout=timeout)
        break # success
    except URLError as err:
        if not isinstance(err.reason, socket.timeout):
           raise # propagate non-timeout errors
else: # all ntries failed 
    raise err # re-raise the last timeout error
# use page here

来源：https://stackoverflow.com/questions/9446387/how-to-retry-urllib2-request-when-fails

标签

python

urllib2

decorator