Question

我有一个问题，想出一个简单的扭曲python代码。根据我在文档中的红色，代码here应该可以在没有未处理错误的情况下工作。

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:

链接延迟的失败不是传递给end（）errback吗？

出于某种原因，我不能在Bula的帖子下面发表评论 我设法通过简单地添加

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)

在每个chainDeferred之后

sync_deferred，其中延迟子项的结果需要“等待”，以便父级可以继续这个结果。

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（示例代码中为d）的错误会传递到d2（示例代码中为l）。这意味着d2 / l最终会出现错误，该错误会传递给第一个也是唯一的错误new_f。 new_f返回其参数，将d2 / l留下错误结果，但不会处理。

如果您的意思是new_f来处理失败，那么您应该让它''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.

Answer 2

在this优秀帖子中，使用chainDeferred时可能出现的潜在问题很少。所以当使用这种方法时，应该小心。

Twisted chainDeferred不按预期工作

2 个答案: