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如何在python run_in_executor方法调用中捕获异常

[英]How to catch exceptions in a python run_in_executor method call

如何在run_in_executor调用的run_long_thing()函数中引发异常? 看起来它正在被吞噬。 我不需要阻塞代码中的函数结果。 它基本上是一个火灾和遗忘功能,但如果有任何问题,我仍需要捕捉异常......

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,如果你调用time.sleep你永远不会结束运行asyncio事件循环所以没有结果检测到。 而不是调用time.sleepdo_it你最好不要做这样的事情

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

现在,run_in_executor的返回是一个未来。 如果您不介意编写异步def并在asyncio循环上使用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

但是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()

那么当你调用run_in_executor时:

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

然后,只要执行程序任务完成,就会调用callback future.result()将包含结果,如果它不是一个例外, future.exception()将返回任何引发的异常

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