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