10:30 AM查看当前时间是否介于 say和之间的最佳方法是什么4:30 PM？
我可以想到以下内容，不确定如何正确：
from datetime import datetime
nw = datetime.now()
hrs = nw.hour;mins = nw.minute;secs = nw.second;
zero = timedelta(seconds = secs+mins*60+hrs*3600)
st = nw - zero # this take me to 0 hours.
time1 = st + timedelta(seconds=10*3600+30*60) # this gives 10:30 AM
time2 = st + timedelta(seconds=16*3600+30*60) # this gives 4:30 PM
if nw>= time1 or nw <= time2:
print "yes, within the interval"
请让我知道这是否是正确的方法，可以写更好的东西吗？
我最初的答案非常具体地集中在提出的问题上，并且不适应跨越午夜的时间范围。由于 6 年后这仍然是公认的答案，因此我在下面合并了@rouble 的答案，以支持午夜。
from datetime import datetime, time
def is_time_between(begin_time, end_time, check_time=None):
# If check time is not given, default to current UTC time
check_time = check_time or datetime.utcnow().time()
if begin_time < end_time:
return check_time >= begin_time and check_time <= end_time
else: # crosses midnight
return check_time >= begin_time or check_time <= end_time
# Original test case from OP
is_time_between(time(10,30), time(16,30))
# Test case when range crosses midnight
is_time_between(time(22,0), time(4,00))
我仍然坚持我在下面的原始评论，即这种逻辑的大多数应用程序可能更适合datetime那些跨越午夜被反映为日期变化的对象。
上述接受的解决方案不适用于通宵时间，这是：
import datetime as dt
def isNowInTimePeriod(startTime, endTime, nowTime):
if startTime < endTime:
return nowTime >= startTime and nowTime <= endTime
else:
#Over midnight:
return nowTime >= startTime or nowTime <= endTime
#normal example:
isNowInTimePeriod(dt.time(13,45), dt.time(21,30), dt.datetime.now().time())
#over midnight example:
isNowInTimePeriod(dt.time(20,30), dt.time(1,30), dt.datetime.now().time())
这是@rouble回答的小例子：
from datetime import datetime
def isNowInTimePeriod(startTime, endTime, nowTime):
if startTime < endTime:
return nowTime >= startTime and nowTime <= endTime
else: #Over midnight
return nowTime >= startTime or nowTime <= endTime
timeStart = '3:00PM'
timeEnd = '11:00AM'
timeNow = '2:59AM'
timeEnd = datetime.strptime(timeEnd, "%I:%M%p")
timeStart = datetime.strptime(timeStart, "%I:%M%p")
timeNow = datetime.strptime(timeNow, "%I:%M%p")
print(isNowInTimePeriod(timeStart, timeEnd, timeNow))
我的 2 美分，它对我有用，而且很容易
while True:
now = datetime.now()
current_time = now.strftime("%H:%M:%S")
start = '19:19:00'
end = '19:19:20'
if current_time > start and current_time < end:
print('in')
print(current_time)
tempo.sleep(1)
else:
print('out')
print(current_time)
看看 py-time-between 包：https ://pypi.org/project/py-time-between/
测试用例：
from datetime import time
from timebetween import is_time_between
def test_is_time_between():
t, s, e = time(0), time(0), time(0)
assert is_time_between(t, s, e)
t, s, e = time(0, 0, 0, 1), time(0), time(0, 0, 0, 2)
assert is_time_between(t, s, e)
t, s, e = time(0, 0, 0, 1), time(0, 0, 0, 1), time(0, 0, 0, 2)
assert is_time_between(t, s, e)
t, s, e = time(0, 0, 0, 2), time(0, 0, 0, 1), time(0, 0, 0, 2)
assert is_time_between(t, s, e)
t, s, e = time(0, 0, 1), time(23, 59, 59), time(0, 0, 2)
assert is_time_between(t, s, e)
t, s, e = time(12, 0, 0), time(23, 59, 59), time(0, 0, 0)
assert is_time_between(t, s, e)
t, s, e = time(23, 59, 57), time(23, 59, 59), time(23, 59, 57)
assert is_time_between(t, s, e)
t, s, e = time(23, 59, 58), time(23, 59, 59), time(23, 59, 57)
assert not is_time_between(t, s, e)
t, s, e = time(22), time(22), time(5, 59, 59)
assert is_time_between(t, s, e)
可以进一步简化：
def isNowInTimePeriod(startTime, endTime, nowTime):
if startTime < endTime:
return startTime <= nowTime <= endTime
else: #Over midnight
return nowTime >= startTime or nowTime <= endTime
我有一个类似的要求，我希望某个任务在每个工作日的上午 9 点到下午 3:30 之间运行。
def somefunction():
cdate = datetime.datetime.strftime(datetime.datetime.now(), "%d-%m-%Y")
if (0 <= time.localtime().tm_wday <= 4) and (datetime.datetime.strptime(cdate + " 09:00:00", "%d-%m-%Y %H:%M:%S") <= datetime.datetime.now() <= datetime.datetime.strptime(cdate + " 15:30:00", "%d-%m-%Y %H:%M:%S")):
<< do something >>
cdate变量以字符串格式获取当前日期。
该条件检查当前工作日是否 >= 0（星期一）和 <= 4（星期五）。它还检查当前时间datetime格式是否 >= 今天的上午 9:00，以及当前时间是否 <= 今天的 15:30。
可能比所选答案更接近 OP 要求的解决方案。它使用日期时间而不是时间作为要检查的对象。它还使用持续时间而不是时间来指定间隔的结束。
from datetime import datetime, time, timedelta
def is_date_within(begin_time, span_time, check_date=None):
"""
Return True if given check_date is within the interval begining
at begin_time with a duration of span_time.
Args:
- begin_time: datetime.time object
- span_time: datetime.timedelta object
- check_date: datetime.datetime object.
If None, default to current UTC date
"""
check_date = check_date or datetime.utcnow()
if check_date.time() >= begin_time:
begin_date = check_date.combine(check_date.date(), begin_time)
else:
begin_date = check_date.combine(check_date.date() - timedelta(days=1),
begin_time)
return begin_date <= check_date <= begin_date + span_time
test_date = datetime(2020, 6, 22, 11, 31)
assert(is_date_within(time(10,30), timedelta(hours=4), test_date) == True)
assert(is_date_within(time(10,30), timedelta(hours=1), test_date) == False)
# Span midnight
assert(is_date_within(time(23,30), timedelta(hours=13), test_date) == True)
assert(is_date_within(time(23,30), timedelta(hours=1), test_date) == False)