python - 在 Python 的调用代码中捕获 __enter__ 中的异常

Question

有没有一种方法可以在上下文管理器的 __enter__ 方法中捕获异常，而无需将整个 with block 包装在 try 中？

class TstContx(object):
    def __enter__(self):
        raise Exception("I'd like to catch this exception")
    def __exit__(self, e_typ, e_val, trcbak):
        pass


with TstContx():
    raise Exception("I don't want to catch this exception")
    pass

我知道我可以在 __enter__() 本身中捕获异常，但是我可以从包含 with 语句的函数访问该错误吗？

从表面上看问题Catching exception in context manager __enter__()似乎是同一件事，但这个问题实际上是关于确保 __exit__ 被调用，而不是将 __enter__ 代码与 with语句附上。

...显然动机应该更清楚。 with 语句正在为完全自动化的过程设置一些日志记录。如果程序在设置日志记录之前失败，那么我不能依赖日志记录来通知我，所以我必须做一些特别的事情。我宁愿在不必添加更多缩进的情况下实现效果，如下所示:

try:
    with TstContx():
        try:
            print "Do something"
        except Exception:
            print "Here's where I would handle exception generated within the body of the with statement"
except Exception:
    print "Here's where I'd handle an exception that occurs in __enter__ (and I suppose also __exit__)"

使用两个 try block 的另一个缺点是 __enter__ 中处理异常的代码出现在 后续主体中处理异常的代码之后用 block 。

Answer 1

可以在__enter__中使用try/except捕获异常，然后将异常实例保存为的实例变量>TstContx 类，允许您在 with block 中访问它:

class TstContx(object):
    def __enter__(self):
        self.exc = None
        try:
            raise Exception("I'd like to catch this exception")
        except Exception as e:
            self.exc = e 
        return self

    def __exit__(self, e_typ, e_val, trcbak):
        pass    

with TstContx() as tst:
    if tst.exc:
        print("We caught an exception: '%s'" % tst.exc)
    raise Exception("I don't want to catch this exception")

输出:

We caught an exception: 'I'd like to catch this exception'.
Traceback (most recent call last):
  File "./torn.py", line 20, in <module>
    raise Exception("I don't want to catch this exception")
Exception: I don't want to catch this exception

但不确定您为什么要这样做......