如何在满足 while 循环条件时跳出 time.sleep()

Question

我正在制作根据交通情况调整的警报。稍微弄乱之后，我决定程序最好每 15 分钟左右从 API 获取数据。但是，我注意到如果在 time.sleep() 处于活动状态时满足 while 循环条件，则在 time.sleep() 的整个持续时间结束之前它不会中断。有没有办法让它在满足条件时中断，即使在 time.sleep() 期间也是如此？提前致谢！

while datetime.datetime.now() < self.wakeuptime: #refers to alarm time

            resp = requests.get(url=url, params=param)
            data = json.loads(resp.content)
            simplifiedtime = datetime.datetime.strptime(data["routes"][0]["legs"][0]["departure_time"]["text"], "%I:%M%p").time()
            #gets suggested departure time from api

            combinedsimplifiedtime=datetime.datetime.combine(today, simplifiedtime)
            self.wakeuptime = combinedsimplifiedtime  - datetime.timedelta(minutes=15)
            #resets wakeuptime value depending on api
            self.timetogo = combinedsimplifiedtime
            print self.wakeuptime
            time.sleep(900)
            #waits 15 minutes and checks again until wakeuptime

Answer 1

您可以使用另一个 while 循环来比从 API 更新的循环更频繁地检查当前时间。而不是 time.sleep(900) 你可以这样做：

start_pause = datetime.datetime.now()
while (datetime.datetime.now()-start_pause).seconds < 900 \
    and datetime.datetime.now() < self.wakeuptime:
    time.sleep(1)

900 秒后仍然只有运行外循环。

如何在满足 while 循环条件时跳出 time.sleep()

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