Question

我已经设置了一个异步循环来创建具有限制的任务列表。

        while current_index < len(numbers):
            if len(tasks) >= NO_CONCURRENT:
                _done, tasks = await asyncio.wait(tasks, return_when=asyncio.FIRST_COMPLETED)
                task: Task = next(iter(_done))
                
                try:
                    task.result()
                except Exception as e:
                    logging.error(f"Error: occured {e}")
                    results.append(task.exception())
                else:
                    results.append(task.result())

它有时会成功处理异常（也记录错误）但有时甚至不处理它只是打印出异常（不记录它）

示例

2021-07-21,03:53:55.082 ERROR {base_events} [default_exception_handler] Task exception was never retrieved
future: <Task finished name='Task-16' coro=<AutoPart.get_all_info() done, defined at /mnt/d/PythonProjects/btyu-part-num-search/copy_main.py:258> exception=PartNumberNotFound('Part number 14111047 is NOT Found.')>
Traceback (most recent call last):
  File "/mnt/d/PythonProjects/btyu-part-num-search/copy_main.py", line 260, in get_all_info
    listing_container_id_num = await self.get_part_number(session)
  File "/mnt/d/PythonProjects/btyu-part-num-search/copy_main.py", line 141, in get_part_number
    raise PartNumberNotFound(self.part_number)
copy_main.PartNumberNotFound: Part number 14112106479456 is NOT Found.

Answer 1

当两个任务碰巧在同一个事件循环迭代中完成时，您的代码无法处理这种情况。在这种情况下， asyncio 返回的 Topic_no Object_Transcript 1 Topic 1 1, 4, 5, 7 2 Topic 2 3, 6, 12, 3, 45 3 Topic 3 5, 9, 12, 14, 3, 9, 7, 2 集将有多个元素，尽管请求“第一个”已完成。在这种情况下，您将仅访问第一个任务的结果，而第二个任务将被销毁，而没有任何人访问其异常，正如警告正确指出的那样。

你应该用一个循环来处理它：

done

另外，请注意变量名前面的首字母 while current_index < len(numbers): if len(tasks) >= NO_CONCURRENT: done, _ = await asyncio.wait(tasks, return_when=asyncio.FIRST_COMPLETED) for task in done: try: task.result() except Exception as e: logging.error(f"Error: occured {e}") results.append(task.exception()) else: results.append(task.result()) 是用于您不使用的值的约定，而不是您使用的值。

异步等待处理异常有时会失败？

1 个答案: