如何運行 function 僅 5 分鍾？

Question

import logging

from a.x.models import X
from a.x.management.commands.syncx \
    import Command as SyncCommand

from a.x.adapter_classes import ADAPTER_CLASSES

LOGGER = logging.getLogger(__name__)


def logger_function(code):
    if not X.objects.filter(code=code).exists():
        X.objects.create(code=code)
        LOGGER.info(f"{X} created")

    args = []
    kwargs = {'x_code': code,
              'class': False,
              'database': 'default'}

    try:
        LOGGER.info(f"Starting syncx command for {code}")
        #or this command needs to be run just 5 minutes for every key
        SyncCommand().handle(*args, **kwargs)
        LOGGER.info(f"There is no error for {code}")

    except Exception as error:
        with open("logger.txt", "a") as file:
            file.write(f"{code}'s error is : {error}")
            LOGGER.info(f"Logging error about {code}\n")


def run():
    for key in ADAPTER_CLASSES.keys():
        #this function needs to be run just 5 minutes for every key
        logger_function(key)

我的 logger_function 需要運行 5 分鍾。 有沒有帶定時器的定時器裝飾器或線程銷毀器？ 我怎樣才能做到這一點。 我的 for 循環正在轉移鍵並發送到記錄器 function，如果除了阻止它的 okey 之外嘗試有任何問題，但如果一切都適合我的 SyncCommand，它可能需要幾個小時，但我只想在前 5 分鍾內記錄錯誤。

Answer 1

有沒有定時器裝飾器

如果您被允許使用外部庫，我建議您查看timeout-decorator 。

Answer 2

    # importing the required module 
import timeit 
  
# code snippet to be executed only once 
mysetup = "from math import sqrt"
  
# code snippet whose execution time is to be measured 
mycode = ''' 
def example(): 
    mylist = [] 
    for x in range(100): 
        mylist.append(sqrt(x)) 
'''
  
# timeit statement 
print timeit.timeit(setup = mysetup, 
                    stmt = mycode, 
                    number = 10000) 

Answer 3

我使用信號庫解決了它

def handler(signum, frame):
    raise Exception(None)

#do stuff func

 for key in ADAPTER_CLASSES.keys():
        signal.signal(signal.SIGALRM, handler)
        signal.alarm(300) #5 min
        logger_function(key)
        signal.alarm(0)