我想在每周二中午 12:00 发送通知我想获得未来星期二中午 12:00 之前的准确时间（以秒为单位）import datetimefrom datetime import date , timedeltatoday = datetime.date.today()Tuesday = datetime.timedelta( (1-today.weekday()) % 7 ) + timedelta(hours=12)seconds_to_call = Tuesday.total_seconds() print(seconds_to_call)# Gives exactly 5 days in seconds 432,000 + timedelta(hours=12) 43200# 以秒为单位给出 5 天 432,000 + timedelta(hours=12) 43200我如何获得星期二 00:00 加上 12 小时 timedelta(hours=12) 43200那么我周二中午 12 点
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30秒到达战场
TA贡献1828条经验 获得超6个赞
from datetime import datetime, timedelta
# get the date and time for now
now = datetime.now()
# get the current day at midnight
today = now.replace(hour=0, minute=0, second=0, microsecond=0)
# get Tuesday at midnight
tues = today + timedelta( (1-today.weekday()) % 7 )
# get the seconds from now until Tuesday at midnight
seconds_to_tues_midnight = (tues - now).total_seconds()
# get the seconds from now until Tuesday at noon
seconds_to_tues_noon = seconds_to_tues_midnight + timedelta(hours=12)/timedelta(seconds=1)
作为函数
from typing import Tuple
def time_to_tuesday(now: datetime) -> Tuple[float, float]:
today = now.replace(hour=0, minute=0, second=0, microsecond=0)
tues = today + timedelta( (1-today.weekday()) % 7 )
midnight = (tues - now).total_seconds()
noon = midnight + timedelta(hours=12)/timedelta(seconds=1)
return midnight, noon
time_to_tuesday(datetime.now())
慕侠2389804
TA贡献1719条经验 获得超6个赞
第一个问题
我想获得未来星期二中午 12:00 之前的准确时间（以秒为单位）
我想您想要从今天开始直到星期二 12:00 的准确时间（以秒为单位）
import datetime
def relative_date(reference, weekday, timevalue):
hour, minute = divmod(timevalue, 1)
minute *= 60
days = reference.weekday()+(reference.weekday() - weekday)
return (reference + datetime.timedelta(days=days)).replace(hour=int(hour), minute=int(minute), second=0, microsecond=0)
today = datetime.datetime.now()
Tuesday = relative_date(today, 1, 12)
seconds_to_call = (Tuesday - today).total_seconds()
print(seconds_to_call)
# OUTPUT: 405778.097769
您必须计算今天和下周二两个日期之间的差异。该函数relative_date计算下周二的确切日期并将小时设置为12:00:00.000。
第二个问题
我想运行程序获取 00:00 之前的秒数，然后获取 12 点之前的秒数 (43,200)
你可以这样做：
import datetime
def relative_date(reference, weekday, timevalue):
hour, minute = divmod(timevalue, 1)
minute *= 60
days = reference.weekday()+(reference.weekday() - weekday)
return (reference + datetime.timedelta(days=days)).replace(hour=int(hour), minute=int(minute), second=0, microsecond=0)
today = datetime.datetime.now()
Tuesday_midnight = relative_date(today, 1, 0)
seconds_to_call = ((Tuesday_midnight - today) + datetime.timedelta(hours=12)).total_seconds()
print(seconds_to_call)
# OUTPUT: 405778.097769
在这种情况下，计算直到午夜的时间是没有用的，您可以直接从中午获取total_seconds，如第一个问题的代码中所示。
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