python - 使用 asyncio.gather 不会引发内部异常

Question

使用 Python 3.7，我试图捕获异常并通过 following an example I found on StackOverflow 重新引发它.虽然该示例确实有效，但它似乎并不适用于所有情况。下面我有两个尝试重新引发异常的异步 Python 脚本。第一个示例有效，它将打印内部和外部异常。

import asyncio

class Foo:
    async def throw_exception(self):
        raise Exception("This is the inner exception")

    async def do_the_thing(self):
        try:
            await self.throw_exception()
        except Exception as e:
            raise Exception("This is the outer exception") from e

async def run():
    await Foo().do_the_thing()

def main():
    loop = asyncio.get_event_loop()
    loop.run_until_complete(run())

if __name__ == "__main__":
    main()

运行它会正确输出以下异常堆栈跟踪:

$ py test.py
Traceback (most recent call last):
  File "test.py", line 9, in do_the_thing
    await self.throw_exception()
  File "test.py", line 5, in throw_exception
    raise Exception("This is the inner exception")
Exception: This is the inner exception

The above exception was the direct cause of the following exception:
Traceback (most recent call last):
  File "test.py", line 21, in <module>
    main()
  File "test.py", line 18, in main
    loop.run_until_complete(run())
  File "C:\Python37\lib\asyncio\base_events.py", line 584, in run_until_complete
    return future.result()
  File "test.py", line 14, in run
    await Foo().do_the_thing()
  File "test.py", line 11, in do_the_thing
    raise Exception("This is the outer exception") from e
Exception: This is the outer exception

但是，在我的下一个 Python 脚本中，我有多个任务要排队，我想从中获取类似的异常堆栈跟踪。本质上，除了上面的堆栈跟踪打印 3 次(以下脚本中的每个任务一次)。上面和下面的脚本之间的唯一区别是 run() 函数。

import asyncio

class Foo:
    async def throw_exception(self):
        raise Exception("This is the inner exception")

    async def do_the_thing(self):
        try:
            await self.throw_exception()
        except Exception as e:
            raise Exception("This is the outer exception") from e

async def run():
    tasks = []

    foo = Foo()

    tasks.append(asyncio.create_task(foo.do_the_thing()))
    tasks.append(asyncio.create_task(foo.do_the_thing()))
    tasks.append(asyncio.create_task(foo.do_the_thing()))

    results = await asyncio.gather(*tasks, return_exceptions=True)

    for result in results:
        if isinstance(result, Exception):
            print(f"Unexpected exception: {result}")

def main():
    loop = asyncio.get_event_loop()
    loop.run_until_complete(run())

if __name__ == "__main__":
    main()

上面的代码片段产生了令人失望的短异常，缺少堆栈跟踪。

$ py test.py
Unexpected exception: This is the outer exception
Unexpected exception: This is the outer exception
Unexpected exception: This is the outer exception

如果我将 return_exceptions 更改为 False，我将打印一次异常和堆栈跟踪，然后停止执行并取消剩余的两个任务。输出与第一个脚本的输出相同。这种方法的缺点是，即使遇到异常我也想继续处理任务，然后在所有任务完成时在最后显示所有异常。

Answer 1

如果您不提供 return_exceptions=True 参数，

asyncio.gather 将在第一个异常处停止，因此您的方法是正确的:您需要收集所有先显示结果和异常，然后显示它们。

要获得您缺少的完整堆栈跟踪，您需要做的不仅仅是“打印”异常。查看 stdlib 中的 traceback 模块，其中包含您需要的所有内容:https://docs.python.org/3/library/traceback.html

您也可以使用 logging.exception，其效果大致相同。