在上下文管理器中处理异常

Question

我有一些代码尝试访问资源，但有时它不可用，并导致异常。我尝试使用 上下文管理器 实现重试引擎，但我无法处理调用者在 __enter__ 上下文中为我的上下文管理器引发的异常。

class retry(object):
    def __init__(self, retries=0):
        self.retries = retries
        self.attempts = 0
    def __enter__(self):
        for _ in range(self.retries):
            try:
                self.attempts += 1
                return self
            except Exception as e:
                err = e
    def __exit__(self, exc_type, exc_val, traceback):
        print 'Attempts', self.attempts

这些只是引发异常（我希望处理）的一些示例

>>> with retry(retries=3):
...     print ok
... 
Attempts 1
Traceback (most recent call last):
  File "<stdin>", line 2, in <module>
NameError: name 'ok' is not defined
>>> 
>>> with retry(retries=3):
...     open('/file')
... 
Attempts 1
Traceback (most recent call last):
  File "<stdin>", line 2, in <module>
IOError: [Errno 2] No such file or directory: '/file'

有没有办法拦截这些异常并在上下文管理器中处理它们？

Answer 1

要处理 __enter__ 方法中的异常，最直接（也不太令人惊讶）的做法是将 with 语句本身包装在 try-except 子句中，并简单地引发异常 -

但是，with 块显然不是设计成这样工作的 - 本身就是 "retriable" - 这里存在一些误解：

def __enter__(self):
    for _ in range(self.retries):
        try:
            self.attempts += 1
            return self
        except Exception as e:
            err = e

一旦你 return self 在那里，__enter__ 运行的上下文不再存在 - 如果 with 块内发生错误，它将自然流动__exit__ 方法。不，无论如何，__exit__ 方法不能使执行流程回到 with 块的开头。

您可能想要更像这样的东西：

class Retrier(object):

    max_retries = 3

    def __init__(self, ...):
         self.retries = 0
         self.acomplished = False

    def __enter__(self):
         return self

    def __exit__(self, exc, value, traceback):
         if not exc:
             self.acomplished = True
             return True
         self.retries += 1
         if self.retries >= self.max_retries:
             return False
         return True

....

x = Retrier()
while not x.acomplished:
    with x:
        ...

Answer 2

引用 __exit__,

If an exception is supplied, and the method wishes to suppress the exception (i.e., prevent it from being propagated), it should return a true value. Otherwise, the exception will be processed normally upon exit from this method.

默认情况下，如果您不 return 一个函数的明确值，Python 将 return None，这是一个虚假值。在您的情况下，__exit__ returns None 这就是允许异常流过 __exit__.

的原因

所以，return 一个真值，像这样

class retry(object):

    def __init__(self, retries=0):
        ...


    def __enter__(self):
        ...

    def __exit__(self, exc_type, exc_val, traceback):
        print 'Attempts', self.attempts
        print exc_type, exc_val
        return True                                   # or any truthy value

with retry(retries=3):
    print ok

输出将是

Attempts 1
<type 'exceptions.NameError'> name 'ok' is not defined

如果你想有重试功能，你可以用装饰器来实现，像这样

def retry(retries=3):
    left = {'retries': retries}

    def decorator(f):
        def inner(*args, **kwargs):
            while left['retries']:
                try:
                    return f(*args, **kwargs)
                except NameError as e:
                    print e
                    left['retries'] -= 1
                    print "Retries Left", left['retries']
            raise Exception("Retried {} times".format(retries))
        return inner
    return decorator


@retry(retries=3)
def func():
    print ok

func()

Answer 3

我觉得这个很简单，其他人似乎想多了。把资源获取代码放在__enter__里，尝试return，而不是self，而是获取资源。在代码中：

def __init__(self, retries):
    ...
    # for demo, let's add a list to store the exceptions caught as well
    self.errors = []

def __enter__(self):
    for _ in range(self.retries):
        try:
            return resource  # replace this with real code
        except Exception as e:
            self.attempts += 1
            self.errors.append(e)

# this needs to return True to suppress propagation, as others have said
def __exit__(self, exc_type, exc_val, traceback):
    print 'Attempts', self.attempts
    for e in self.errors:
        print e  # as demo, print them out for good measure!
    return True

现在试试看：

>>> with retry(retries=3) as resource:
...     # if resource is successfully fetched, you can access it as `resource`;
...     # if fetching failed, `resource` will be None
...     print 'I get', resource
I get None
Attempts 3
name 'resource' is not defined
name 'resource' is not defined
name 'resource' is not defined

在上下文管理器中处理异常

Handling exceptions inside context managers

python

exception

contextmanager