突破或退出“with"陈述?

问题描述

我只想在某些条件下退出 with 语句:

I'd just like to exit out of a with statement under certain conditions:

with open(path) as f:
    print 'before condition'
    if <condition>: break #syntax error!
    print 'after condition'

当然，以上是行不通的.有没有办法做到这一点?(我知道我可以反转条件:if not <condition>: print 'after condition' - 有没有像上面那样的方式?)

Of course, the above doesn't work. Is there a way to do this? (I know that I can invert the condition: if not <condition>: print 'after condition' -- any way that is like above?)

推荐答案

最好的方法是将其封装在一个函数中并使用return:

The best way would be to encapsulate it in a function and use return:

def do_it():
    with open(path) as f:
        print 'before condition'
        if <condition>:
            return
        print 'after condition'

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