Python中的条件语句 [英] Conditional with statement in Python

问题描述

有没有办法用 with 语句开始一段代码，但有条件?

Is there a way to begin a block of code with a with statement, but conditionally?

类似:

if needs_with():
    with get_stuff() as gs:

# do nearly the same large block of stuff,
# involving gs or not, depending on needs_with()

为了澄清，一种情况是在 with 语句中包含一个块，而另一种可能性是同一个块，但没有被封装(即，好像它没有缩进)

To clarify, one scenario would have a block encased in the with statement, while another possibility would be the same block, but not encased (i.e., as if it wasn't indented)

初始实验当然会出现缩进错误..

Initial experiments of course give indentation errors..

推荐答案

如果你想避免重复代码并使用 3.7 之前的 Python 版本(当引入 contextlib.nullcontext 时)或即使是 3.3(引入 contextlib.ExitStack 时)，您也可以执行以下操作:

If you want to avoid duplicating code and are using a version of Python prior to 3.7 (when contextlib.nullcontext was introduced) or even 3.3 (when contextlib.ExitStack was introduced), you could do something like:

class dummy_context_mgr():
    def __enter__(self):
        return None
    def __exit__(self, exc_type, exc_value, traceback):
        return False

或:

import contextlib

@contextlib.contextmanager
def dummy_context_mgr():
    yield None

然后将其用作:

with get_stuff() if needs_with() else dummy_context_mgr() as gs:
   # do stuff involving gs or not

您也可以根据 needs_with() 让 get_stuff() 返回不同的内容.

You alternatively could make get_stuff() return different things based on needs_with().

(请参阅 Mike 的回答 或 Daniel 的回答 了解您可以在以后的版本中做什么.)

(See Mike's answer or Daniel's answer for what you can do in later versions.)

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