使用asyncio.gather不会引发内部异常

Question

使用Python 3.7，我试图捕获一个异常并通过following an example I found on StackOverflow重新提升它。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。第一个例子有效，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()

运行此命令将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，我排队，我想从中获得类似的异常堆栈跟踪。基本上，除了上面的堆栈跟踪要打印3次（以下脚本中的每个任务一次）。上面和下面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暂异常，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，我将打印出异常和堆栈跟踪一次，然后执行停止，剩下的两个任务被取消。输出与第一个脚本的输出相同。这种方法的缺点是，我希望即使遇到异常也要继续处理任务，然后在所有任务完成时显示所有异常。

Answer 1

如果你不提供asyncio.gather参数，return_exceptions=True将停在第一个例外，所以你的方法是正确的：你需要先收集所有结果和异常，然后显示它们。

要获得缺少的完整堆栈跟踪，您需要做的不仅仅是“打印”异常。看看stdlib中的traceback模块，它具备您所需要的一切：https://docs.python.org/3/library/traceback.html

您也可以使用logging.exception，它或多或少会相同。