上下文管理器的动态可迭代链接到单个with语句

我有一堆要链接的上下文管理器.乍一看,contextlib.nested看起来很合适.但是,该方法在文档中被标记为已弃用,该文档还指出,最新的with语句可直接允许这样做:

Deprecated since version 2.7: The with-statement now supports this
functionality directly (without the confusing error prone quirks).

但是我无法让Python 3.4.3使用上下文管理器的动态迭代:

class Foo():
    def __enter__(self):
        print('entering:', self.name)
        return self
    def __exit__(self, *_):
        pass
    def __init__(self, name):
        self.name = name

foo = Foo('foo')
bar = Foo('bar')

是否链接:

from itertools import chain
m = chain([foo], [bar])
with m:
     pass

Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
AttributeError: __exit__
m = [foo, bar]

直接提供清单:

with m:
     pass

Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
AttributeError: __exit__

或拆箱:

with (*m):
    pass

  File "<stdin>", line 1
SyntaxError: can use starred expression only as assignment target

那么,如何正确地在with语句中正确链接动态数量的上下文管理器?

解决方法:

您误解了那条线. with语句需要多个上下文管理器,以逗号分隔,但不是可迭代的:

with foo, bar:

作品.

如果需要支持一组动态上下文管理器,请使用contextlib.ExitStack() object

from contextlib import ExitStack

with ExitStack() as stack:
    for cm in (foo, bar):
        stack.enter_context(cm)
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