How to solve my deadlineexceedederror for a GAE cron task? - google-app-engine

Question

I'm getting a deadlineexceedederror when trying to create a kml file by a cron that starts a task. The error message is
2012-01-30 16:07:47.902 /createkml/ 500 5012ms 0kb AppEngine-Google; (+http://code.google.com/appengine)
0.1.0.2 - - [30/Jan/2012:10:07:47 -0800] "POST /createkml/ HTTP/1.1" 500 0 "http://www.koolbusiness.com/createkmltask/" "AppEngine-Google; (+http://code.google.com/appengine)" "www.koolbusiness.com" ms=5013 cpu_ms=3305 api_cpu_ms=295 cpm_usd=0.091855 queue_name=default task_name=7177535986681131286 instance=00c61b117cf890b99ca52a6b9e7c5f048e72
I 2012-01-30 16:07:42.959
creating file
E 2012-01-30 16:07:47.853
ApplicationError: 5
Traceback (most recent call last):
File "/base/python27_runtime/python27_lib/versions/third_party/webapp2-2.3/webapp2.py", line 1511, in __call__
rv = self.handle_exception(request, response, e)
File "/base/python27_runtime/python27_lib/versions/third_party/webapp2-2.3/webapp2.py", line 1505, in __call__
rv = self.router.dispatch(request, response)
File "/base/python27_runtime/python27_lib/versions/third_party/webapp2-2.3/webapp2.py", line 1253, in default_dispatcher
return route.handler_adapter(request, response)
File "/base/python27_runtime/python27_lib/versions/third_party/webapp2-2.3/webapp2.py", line 1077, in __call__
return handler.dispatch()
File "/base/python27_runtime/python27_lib/versions/third_party/webapp2-2.3/webapp2.py", line 547, in dispatch
return self.handle_exception(e, self.app.debug)
File "/base/python27_runtime/python27_lib/versions/third_party/webapp2-2.3/webapp2.py", line 545, in dispatch
return method(*args, **kwargs)
File "/base/data/home/apps/s~montaoproject/gralumo.356457066048686089/main.py", line 1575, in post
result = urlfetch.fetch(url)
File "/base/python27_runtime/python27_lib/versions/1/google/appengine/api/urlfetch.py", line 263, in fetch
return rpc.get_result()
File "/base/python27_runtime/python27_lib/versions/1/google/appengine/api/apiproxy_stub_map.py", line 592, in get_result
return self.__get_result_hook(self)
File "/base/python27_runtime/python27_lib/versions/1/google/appengine/api/urlfetch.py", line 371, in _get_fetch_result
raise DeadlineExceededError(str(err))
DeadlineExceededError: ApplicationError: 5
My code is
class CreateKMLTask(webapp2.RequestHandler):
def get(self):
logging.info('creating KML task')
taskqueue.add(url=r'/createkml/')
self.redirect('/')
class CreateKMLHandler(webapp2.RequestHandler):
def post(self):
# Create the file
logging.info('creating file')
file_name = \
files.blobstore.create(mime_type='application/octet-stream')
url = 'http://montaoproject.appspot.com/list.kml'
result = urlfetch.fetch(url)
if not result.content:
return
# Open the file and write to it
logging.info('opening file')
with files.open(file_name, 'a') as f:
f.write(result.content)
# Finalize the file. Do this before attempting to read it.
files.finalize(file_name)
# Get the file's blob key
logging.info('getting blob key of file')
blob_key = files.blobstore.get_blob_key(file_name)
logging.info('new blob key:'+str(blob_key))
im = Image.get_by_id(4468185)
im.primary_image = blob_key
im.put()
logging.info('new image key:'+str(im.primary_image))
logging.info('finished KML handling and put image')
webapp2.Route(r'/createkml/', handler=CreateKMLHandler,
name='createkml'),
webapp2.Route(r'/createkmltask/', handler=CreateKMLTask,
name='createkmltask')
My cron.yaml:
cron:
- description: daily mailout
url: /report/daily
schedule: every 24 hours
- description: daily mailout
url: /report/montaodaily
schedule: every 24 hours
- description: refresh kml
url: /createkmltask/
schedule: every 24 hours
What I'm trying to do is refresh to KML file to a new blob that has the same File ID as before so the call will be the same. What can I do to resolve my timeout problem? I think it should work.
Thanks

Answer 1

This timeout is fetching the url; you can see this in the stack trace. The fetch function defaults to a 5-second timeout.
The URL Fetch docs say you can increase the timeout by passing deadline as an additional argument. For example, deadline=60 would give you a 60-second timeout.
So, try result = urlfetch.fetch(url, deadline=60)