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Retry requests on server error ****************************** If lazr.restfulclient talks to a server that sends out a server-side error with status codes 502 or 503, the client will wait a few seconds and try the request again. Eventually it will give up and escalate the error code in the form of an exception. To test this, let's simulate a lazr.restful server prone to transient errors using a WSGI application. >>> import pkg_resources >>> wadl_string = pkg_resources.resource_string( ... 'wadllib.tests.data', 'launchpad-wadl.xml') >>> representations = { 'application/vnd.sun.wadl+xml' : wadl_string, ... 'application/json' : '{}' } This application will cause one request to fail for every item in its BROKEN_RESPONSES list. >>> BROKEN_RESPONSES = [] >>> def broken_application(environ, start_response): ... if len(BROKEN_RESPONSES) > 0: ... start_response(str(BROKEN_RESPONSES.pop()), ... [('Content-type', 'text/plain')]) ... return ["Sorry, I'm still broken."] ... else: ... media_type = environ['HTTP_ACCEPT'] ... content = representations[media_type] ... start_response( ... '200', [('Content-type', media_type)]) ... return [content] >>> def make_broken_application(): ... return broken_application >>> import wsgi_intercept >>> wsgi_intercept.add_wsgi_intercept( ... 'api.launchpad.dev', 80, make_broken_application) >>> BROKEN_RESPONSES = [] >>> from wsgi_intercept.httplib2_intercept import install >>> install() Here's a fake implementation of time.sleep() so that this test doesn't take a really long time to run, and so we can visualize sleep() being called as lazr.restfulclient retries over and over again. >>> def fake_sleep(time): ... print "sleep(%s) called" % time >>> import lazr.restfulclient._browser >>> old_sleep = lazr.restfulclient._browser.sleep >>> lazr.restfulclient._browser.sleep = fake_sleep As it starts out, the application isn't broken at all. >>> from lazr.restfulclient.resource import ServiceRoot >>> client = ServiceRoot(None, "http://api.launchpad.dev/") Let's queue up one broken response. The client will sleep once and try again. >>> BROKEN_RESPONSES = [502] >>> client = ServiceRoot(None, "http://api.launchpad.dev/") sleep(0) called Now the application will fail six times and then start working. >>> BROKEN_RESPONSES = [502, 503, 502, 503, 502, 503] >>> client = ServiceRoot(None, "http://api.launchpad.dev/") sleep(0) called sleep(1) called sleep(2) called sleep(4) called sleep(8) called sleep(16) called Now the application will fail seven times and then start working. But the client will give up before then--it will only retry the request six times. >>> BROKEN_RESPONSES = [502, 503, 502, 503, 502, 503, 502] >>> client = ServiceRoot(None, "http://api.launchpad.dev/") Traceback (most recent call last): ... ServerError: HTTP Error 502: ... By increasing the 'max_retries' constructor argument, we can make the application try more than six times, and eventually succeed. >>> BROKEN_RESPONSES = [502, 503, 502, 503, 502, 503, 502] >>> client = ServiceRoot(None, "http://api.launchpad.dev/", ... max_retries=10) sleep(0) called sleep(1) called sleep(2) called sleep(4) called sleep(8) called sleep(16) called sleep(32) called Now the application will fail once and then give a 400 error. The client will not retry in hopes that the 400 error will go away--400 is a client error. >>> BROKEN_RESPONSES = [502, 400] >>> client = ServiceRoot(None, "http://api.launchpad.dev/") Traceback (most recent call last): ... BadRequest: HTTP Error 400: ... Teardown. >>> wsgi_intercept.remove_wsgi_intercept("api.launchpad.dev", 80) >>> lazr.restfulclient._browser.sleep = old_sleep