@retry_on_timeout(retries=10, secs=0.2)
def some_idempotent_function(): # do stuff
If you do not give retries or secs it defaults to 3 retries with 1 sec
starting delay. The delay is exponential, it doubles after each retry.
2009/8/25 Devel63 <danstic...@gmail.com>
>
> Can you give an example as to how this is used? I understand the
> purpose, I'm just a little hazy on the calling syntax.
>
> On Aug 24, 5:25 am, Alkis Evlogimenos ('Αλκης Ευλογημένος)
> <evlogime...@gmail.com> wrote:
> > You can make it into a decorator which will make it easier for your
> > functions to code. I use this:
> > def retry_on_timeout(retries=3, secs=1):
> > """A decorator to retry a given function performing db operations."""
> > def _decorator(func):
> > def _wrapper(*args, **kwds):
> > tries = 0
> > while True:
> > try:
> > tries += 1
> > return func(*args, **kwds)
> > except db.Timeout, e:
> > logging.debug(e)
> > if tries > retries:
> > raise e
> > else:
> > wait_secs = secs * tries ** 2
> > logging.warning("Retrying function %r in %d secs" % (func,
> > wait_secs))
> > time.sleep(wait_secs)
> > return _wrapper
> > return _decorator
> >
> >
> >
> > On Mon, Aug 24, 2009 at 1:53 PM, Bemmu <bemmu....@gmail.com> wrote:
> >
> > > I decided to finally do something about the Timeout exceptions
> > > littering my log.
> >
> > > I read somewhere on this forum that I am supposed to code around data
> > > store accesses to try things out several times in case of timeouts. Is
> > > this still necessary? Why won't the methods just do that internally?
> >
> > > This is my first attempt to handle a timeout situation, is there any
> > > nicer way to code this?
> >
> > > retries = 3
> > > retry = True
> > > while retry:
> > > try:
> > > retry = False
> > > recipient.put()
> > > except:
> > > retries -= 1
> > > if retries > 0:
> > > retry = True
> > > logging.info("recipient.put() failed,
> > > retrying")
> > > else:
> > > logging.error("failed even after trying
> > > recipient.put() three
> > > times")
> >
> > --
> >
> > Alkis
> >
>
--
Alkis
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