Python - 在给定时间启动一个函数 [英] Python - Start a Function at Given Time

问题描述

如何在给定时间在 Python 中运行函数?

How can I run a function in Python, at a given time?

例如:

run_it_at(func, '2012-07-17 15:50:00')

它将在 2012-07-17 15:50:00 运行 func 函数.

and it will run the function func at 2012-07-17 15:50:00.

我尝试了 sched.scheduler，但没有启动我的功能.

I tried the sched.scheduler, but it didn't start my function.

import time as time_module
scheduler = sched.scheduler(time_module.time, time_module.sleep)
t = time_module.strptime('2012-07-17 15:50:00', '%Y-%m-%d %H:%M:%S')
t = time_module.mktime(t)
scheduler_e = scheduler.enterabs(t, 1, self.update, ())

我能做什么?

推荐答案

从 阅读文档http://docs.python.org/py3k/library/sched.html:

因此，我们需要计算出延迟(以秒为单位)...

Going from that we need to work out a delay (in seconds)...

from datetime import datetime
now = datetime.now()

然后使用datetime.strptime解析'2012-07-17 15:50:00'(格式字符串留给你)

Then use datetime.strptime to parse '2012-07-17 15:50:00' (I'll leave the format string to you)

# I'm just creating a datetime in 3 hours... (you'd use output from above)
from datetime import timedelta
run_at = now + timedelta(hours=3)
delay = (run_at - now).total_seconds()

然后您可以使用 delay 传递给 threading.Timer 实例，例如:

You can then use delay to pass into a threading.Timer instance, eg:

threading.Timer(delay, self.update).start()

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