backoff if exception argument == ... about backoff

[luckydonald]

Like the @on_predicate lambda

[luckydonald]

Actual really dirty workaround:

class DoRetryException(BaseException):
    """
    Used for backoff strategy
    """
    pass
# end class DoRetryException


def should_backoff(e):
    """
    Used for backoff strategy
    """
    if e.error_code == 429 or "too many requests" in e.description or "retry later" in e.description:
        raise DoRetryException()
    # end if
# end def



class DocumentMessage():
    # ...
    # ...

    @backoff.on_exception(backoff.expo, DoRetryException, max_tries=10, jitter=None)
    def send(self, sender: PytgbotApiBot, receiver, reply_id)->PytgbotApiMessage:
        try:
            return sender.send_document(
                receiver, self.file, caption=self.caption, reply_to_message_id=reply_id, reply_markup=self.reply_markup,
                disable_notification=self.disable_notification
            )
        except TgApiServerException as e:
            should_backoff(e)  # checks if it should raise an DoRetryException
            raise  # else it just raises as usual
        # end try
    # end def send

[bgreen-litl]

I think this is basically the same as #8 ... I'm still thinking about the best way to accommodate it. Does the keyword arg solution I put there look like it would work for you?

[bgreen-litl]

The idea is to add a keyword arg which takes a function which tests the raised exception.

    def is_fatal(e):
        return e.error_number == 42


    @backoff.on_exception(backoff.expo, SomeException,
                          giveup=is_fatal)
    def foobar():
        error_number = random.choice([42, 4458])
        raise SomeException(error_number)

How does this look to you?

[bgreen-litl]

@luckydonald This PR adds a giveup kwarg to on_exception: #13

[luckydonald]

Looks like the missing part :D

[luckydonald]

Should giveup(e) be checked for raising exceptions itself, so another backoff round starts? Or should that just fail?

[bgreen-litl]

If there's an exception raised in your giveup function I believe that should be allowed to escape.

[luckydonald]

Yep, probably. Else errors there would be unnoticed.

[luckydonald]

When will there be a new pip package? or did I miss that already?

[bgreen-litl]

Yes, this will be released and distributed on pypi as part of 1.3

[bgreen-litl]

backoff==1.3.1 is now released

[luckydonald]

Great!

[luckydonald]

#11 (comment)
In your code you say @backoff.on_exception(...): where the : is wrong.

Also could you please syntax highlight to your code (here and readme) as python by using ```python?