python - 安排Python程序在给定时间段内休眠

Question

while True:
      now = datetime.datetime.now();
      if now.hour >= 22 and now.hour < 3:
         print "sleep"
         sleep_at = datetime.datetime.combine(datetime.date.today(),datetime.time(3))
         sleep_til = (sleep_at - now).seconds
         print sleep_til
         time.sleep(sleep_til)
         print "wake"
      else:
         print "break"
         break

这段代码应该让我的整个程序在晚上 10 点进入休眠状态并在凌晨 3 点醒来。我的问题是..这会起作用吗？我尝试运行它..但我无法更改我的系统/计算机时间..所以我无法检查。我发布这个问题的原因是因为我的编码使用 datetime.date.tday 和 datetime.datetime 调用当前日期..

再一次..我希望我的程序在晚上 10 点之前运行，并在晚上 10 点到凌晨 3 点之间休眠，并在凌晨 3 点之后重新运行。

有人可以检查这是否是正确的方法吗？

Answer 1

考虑(为了清晰起见，更加详细):

import time, datetime

# Create time bounds -- program should run between RUN_LB and RUN_UB
RUN_LB = datetime.time(hour=22)     # 10pm
RUN_UB = datetime.time(hour=3)      #  3am

# Helper function to determine whether we should be currently running
def should_run():
    # Get the current time
    ct = datetime.datetime.now().time()
    # Compare current time to run bounds
    lbok = RUN_LB <= ct
    ubok = RUN_UB >= ct
    # If the bounds wrap the 24-hour day, use a different check logic
    if RUN_LB > RUN_UB:
        return lbok or ubok
    else:
        return lbok and ubok


# Helper function to determine how far from now RUN_LB is
def get_wait_secs():
    # Get the current datetime
    cd = datetime.datetime.now()
    # Create a datetime with *today's* RUN_LB
    ld = datetime.datetime.combine(datetime.date.today(), RUN_LB)
    # Create a timedelta for the time until *today's* RUN_LB
    td = ld - cd
    # Ignore td days (may be negative), return td.seconds (always positive)
    return td.seconds


while True:
    if should_run():
        print("--do something--")
    else:
        wait_secs = get_wait_secs()
        print("Sleeping for %d seconds..." % wait_secs)
        time.sleep(wait_secs)

但我确实同意 sleep 并不是延迟程序开始的最佳方法。您可以查看 Windows 上的任务计划程序或 Linux 上的 cron。