如何计算从现在开始加上n天的总秒数

Question

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 datetime

from datetime import date , timedelta

today = 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

Answer 1

from datetime import datetime, timedelta# get the date and time for nownow = datetime.now()# get the current day at midnighttoday = now.replace(hour=0, minute=0, second=0, microsecond=0)# get Tuesday at midnighttues = today + timedelta( (1-today.weekday()) % 7 )# get the seconds from now until Tuesday at midnightseconds_to_tues_midnight = (tues - now).total_seconds()# get the seconds from now until Tuesday at noonseconds_to_tues_noon = seconds_to_tues_midnight + timedelta(hours=12)/timedelta(seconds=1)作为函数from typing import Tupledef 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, noontime_to_tuesday(datetime.now())

Answer 2

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 datetimedef 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 datetimedef 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。