Question

我试图找出有人会在10亿秒之前变老的时候。用户在他们出生时输入。然后将这些值转换为秒，然后再添加10亿秒并转换回日期。但是，当我输入某些日期时，python似乎搞得一团糟。这样的例子是1993/11/05 00:00:00，其中python说用户将在第0个月转向。注意我不能使用if / else或datetime。

继承我的代码：

YEARSEC=(12*30*24*3600) 
MONTHSEC=(3600*24*30) 
DAYSEC=(24*3600)
HOURSEC=3600
MINUTESEC=60

year=int(input("Please enter the year in which you were born: ")) 
month=int(input("Please enter the month you were born: "))    
day=int(input("Please enter the day you were born: "))
hour=int(input("Please enter the hour you were born: "))
minute=int(input("Please enter the minute you were born: "))
second=int(input("Please enter the second you were born: "))

year_calc=(year*YEARSEC)  
month_calc=(month*MONTHSEC)
day_calc=(day*DAYSEC)
hour_calc=(hour*HOURSEC)
minute_calc=(minute*MINUTESEC)
s=(1000000000+year_calc+month_calc+day_calc+hour_calc+minute_calc+second) 


year_num=int((s/YEARSEC)) 
s=(s-(year_num*YEARSEC))


month_num=int((s/MONTHSEC))
s=(s-(month_num*MONTHSEC))


day_num=int((s/DAYSEC))
s=(s-(DAYSEC*day_num))

hour_num=int((s/HOURSEC))
s=(s-(HOURSEC*hour_num))

minute_num=int((s/MINUTESEC))
s=(s-(MINUTESEC*minute_num))




print("You will turn 1 000 000 000 seconds old on: %04d/%02d/%02d %02d:%02d:%02d" %(year_num,month_num,day_num,hour_num,minute_num,s))

Answer 1

时间计算很棘手。例如，几个月都没有30天。小时，分钟和秒从0开始编号，但是日期和月份从1开始编号，在计算中创建一个错误（提示，请求月份，然后减去一个，执行所有计算，然后添加一次再显示时）。你也没有考虑到闰年。

最好使用内置工具，如果只是检查你最终的作业答案，虽然看起来老师说要假设30天的月份; ^）

>>> import datetime
>>> birthday = datetime.datetime(1993,11,05,0,0,0)
>>> billion = birthday + datetime.timedelta(seconds=1000000000)
>>> billion.ctime()
'Mon Jul 14 01:46:40 2025'

Answer 2

虽然我没有全部测试，但我认为你不能得到12月和30日。你应该加1到day_num和month_num导致月和日从1开始而不是0。

print("You will turn 1 000 000 000 seconds old on: %04d/%02d/%02d %02d:%02d:%02d" %(year_num,month_num+1,day_num+1,hour_num,minute_num,s))

关于简单的python日期计算

2 个答案: