python - Twisted chainDeferred 未按预期工作

Question

我在弄清楚有点简单的扭曲 python 代码时遇到问题。根据我在文档中的红色内容，代码 here应该可以正常工作，没有未处理的错误。

我明白了:

HELLO!
HANDLED!
HANDLED 2!
Unhandled error in Deferred:
Unhandled Error
Traceback (most recent call last):
  File "package_tester.py", line 31, in <module>
    a().callback(2)
  File "/usr/local/lib/python2.7/dist-packages/twisted/internet/defer.py", line 368, in callback
    self._startRunCallbacks(result)
  File "/usr/local/lib/python2.7/dist-packages/twisted/internet/defer.py", line 464, in _startRunCallbacks
    self._runCallbacks()
--- <exception caught here> ---
  File "/usr/local/lib/python2.7/dist-packages/twisted/internet/defer.py", line 551, in _runCallbacks
    current.result = callback(current.result, *args, **kw)
  File "package_tester.py", line 5, in c
    raise Exception()
exceptions.Exception:

链式延迟的失败不是传递给 end() errback 吗？

由于某种原因，我无法在 Bula 的帖子下发表评论
我通过简单地添加来绕过“unexpected1.py”的行为

@defer.inlineCallbacks
def sync_deferred(self, result, deferred):
        """
        Wait for a deferred to end before continuing.
        @param deferred: deferred which will be waited to finish so the chain
        can continue.
        @return: result from the deferred.
        """
        r = yield deferred
        defer.returnValue(r)

sync_deferred 在每个 chainDeferred 之后，其中子延迟的结果需要“等待”，以便父级可以继续处理此结果。

Answer 1

来自the documentation of Deferred.chainDeferred :

When you chain a deferred d2 to another deferred d1 with d1.chainDeferred(d2),
you are making d2 participate in the callback chain of d1. Thus any event that
fires d1 will also fire d2.

您提供给d1(示例代码中的d)的错误将传递给d2(l > 在您的示例代码中)。这意味着 d2/l 最终会出现错误，该错误会传递给其第一个也是唯一的 errback new_f。 new_f 返回其参数，使 d2/l 带有错误结果，该结果未得到处理。

如果您的意思是 new_f 来处理失败，那么您应该使其“不”返回失败。来自 the Deferred howto :

If the errback does returns a Failure or raise an exception, then that is passed
to the next errback, and so on.