如果我处于一天中的特定时间段,如何检查python? [英] How to check in python if I'm in certain range of times of the day?
问题描述
我想检查python当前时间是否在两个端点之间(例如,上午8:30和下午3:00),与实际日期无关.就像我一样,我不在乎整个日期是多少.只是一个小时.当我使用 strptime
创建 datetime
对象以指定时间时,它放入了一个虚拟日期(我认为是1900年?),这不是我想要的.我可以使用笨拙的布尔表达式,例如(hour == 8 and minutes> = 30)或(9< = hour< 15)
,但这似乎并不优雅.什么是最简单,最Python的方法来完成此任务?
I want to check in python if the current time is between two endpoints (say, 8:30 a.m. and 3:00 p.m.), irrespective of the actual date. As in, I don't care what the full date is; just the hour. When I created datetime
objects using strptime
to specify a time, it threw in a dummy date (I think the year 1900?) which is not what I want. I could use a clumsy boolean expression like (hour == 8 and minute >= 30) or (9 <= hour < 15)
but that doesn't seem very elegant. What's the easiest and most pythonic way to accomplish this?
再延伸一点,我真正想要的是可以告诉我是否在几个小时的范围内,又是一个工作日.当然,我可以使用 0< = weekday()< = 4
来破解,但是可能会有更好的方法.
Extending a bit further, what I'd really like is something that will tell me if it's between that range of hours, and that it's a weekday. Of course I can just use 0 <= weekday() <= 4
to hack this, but there might be a better way.
推荐答案
datetime
对象具有称为 time()
的方法,该方法返回 time
对象(无日期信息).然后,您可以使用普通的<
或>
运算符比较 time
对象.
datetime
objects have a method called time()
which returns a time
object (with no date information). You can then compare the time
objects using the normal <
or >
operators.
import datetime
import time
timestamp = datetime.datetime.now().time() # Throw away the date information
time.sleep(1)
print (datetime.datetime.now().time() > timestamp) # >>> True (unless you ran this one second before midnight!)
# Or check if a time is between two other times
start = datetime.time(8, 30)
end = datetime.time(15)
print (start <= timestamp <= end) # >>> depends on what time it is
如果您还想检查工作日,建议的代码可能是解决问题的最有效方法,但是在那种情况下,您可能不想扔掉原始的 datetime
对象.
now = datetime.datetime.now()
if 0 <= now.weekday() <= 4:
print ("It's a weekday!")
print (start <= now.time() <= end) # with start & end defined as above
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