如何在 python run_in_executor 方法调用中捕获异常

How to catch exceptions in a python run_in_executor method call

如何在使用 run_in_executor 调用的 run_long_thing() 函数中引发异常? 看起来像是被吞噬了一样。我不需要阻塞代码中函数的结果。它基本上是一个即发即忘的功能,但如果有任何异常,我仍然需要捕获异常...

import asyncio
import time


def fire_and_forget(task, *args, **kwargs):
    loop = asyncio.get_event_loop()
    if callable(task):
        #if threadpoolworker is set to None,
        #the max_workers will default to the number of processors on the machine, multiplied by 5
        return loop.run_in_executor(None, task, *args, **kwargs)
    else:    
        raise TypeError('Task must be a callable.')


async def run_long_thing(sleep):
    print("Doing long thing... {:}".format(sleep))
    time.sleep(sleep)
    print("Done doing long thing. {:}".format(sleep))
    raise Exception("sh*t happens")


def do_it():
    print("Starting my main thing...")
    print("Calling my long thing...")
    for i in range(0,10,1):
        try:
            fire_and_forget(run_long_thing, i)
            print(i)
            print("Pom pi dom...")
            time.sleep(0.1)
            print("POOOOM Pom pi dom...")
        except:
            print("can i see the sh*t?")

do_it()

首先,如果您调用 time.sleep,您将永远不会结束 运行 asyncio 事件循环,因此不会检测到任何结果。不要在 do_it 中调用 time.sleep 你最好做类似

的事情
asyncio.get_event_loop().run_until_complete(asyncio.sleep(0.1))

现在,来自 run_in_executor 的 return 是一个未来。如果您不介意编写异步定义并在 asyncio 循环中使用 create_task,您可以执行类似

的操作
async def run_long_thing(thing, *args):
    try: await asyncio.get_event_loop().run_in_executor(None, thing, *args)
    except:
        #do stuff

但更符合您当前的代码,您可以附加异常回调

def callback(future):
if future.exception(): #your long thing had an exception
        # do something with future.exception()

然后当你调用 run_in_executor:

future = asyncio.get_event_loop().run_in_executor(None, fun, *args)
future.add_done_callback(callback)

然后 callback 将在您的执行程序任务完成时被调用。如果没有异常,future.result() 将包含结果,future.exception() 将返回任何引发的异常