How to make months not exceed 12

Question

I am working on a small project where the code will approximately calculate what the date will be when they are one billion seconds old. I am done but I have a problem. If a user enters "5" as month or higher, then the month will exceed 12. Same for the date, it will go over 31 days if the user enters "24" or higher. How do I make it so that it will not go above "12" for months and "31" for days. Only problem is the month and the day, the year is working fine. Thanks!

When running the code, a sample you can use is, "10" as month, "24" as day and "2020" for year.

Code:

month = int(input("What Month Were You Born In: "))
day = int(input("What Day Was It: "))
year = int(input("What Year Was It: "))

sum1 = day + 7
sum2 = month + 8
sum3 = year + 32

print("You will be a billion seconds old approximately around, " + str(sum1) + "/" + str(sum2) + "/" + str(sum3))

Does this answer your question? [Convert timedelta to total seconds](https://stackoverflow.com/questions/5522031/convert-timedelta-to-total-seconds) — Joe Ferndz, Jan 11 '21 at 03:24

import huh · Answer 1 · 2021-11-04T17:44:17.450

While you could do this using if statements, it's cleaner to use the datetime built-in library, which natively handles dates.

import datetime # imports it so it can be used

month = int(input("What Month Were You Born In: ")) #your code
day = int(input("What Day Was It: "))
year = int(input("What Year Was It: "))

born = datetime.datetime(year, month, day)#creates a datetime object which contains a date. Time defaults to midnight.

output = born + datetime.timedelta(seconds = 1_000_000_000)#defines and applies a difference of 1 billion seconds to the date

print(output.strftime("%m/%d/%Y")) #prints month/day/year

If you do want to do it without datetime, you can use this:

month = int(input("What Month Were You Born In: "))
day = int(input("What Day Was It: "))
year = int(input("What Year Was It: "))

sum1 = day + 7
sum2 = month + 8
sum3 = year + 32

if sum1 > 31: # checks if day is too high, and if it is, corrects that and addr 1 to the month to account for that.
    sum1 -= 31
    sum2 += 1
    
if sum2 > 12: #does the same thing for month
    sum2 -= 12
    sum3 += 1
    
print("You will be a billion seconds old approximately around, " + str(sum1) + "/" + str(sum2) + "/" + str(sum3))

Note that this option is less precise than the datetime option, since it doesn't take into account leapyears or various lengths of months.

Thanks, I will be sure to try this out as I am more familiar with that. (a beginner) — koscile, Jan 11 '21 at 04:23
You're welcome. Since you're new here, please don't forget to mark the answer accepted which helped most in solving the problem. — import huh, Jan 11 '21 at 22:40

score 2 · Answer 2 · answered Jan 11 '21 at 02:30

You should use the datetime module to correctly handle dates which will handle issues like this.

timedelta will allow you to add, subtract etc on datetime objects.

from datetime import datetime, timedelta

month = 6
day = 24
year = 1990

born = datetime(month=month, year=year, day=day)

billion = born + timedelta(seconds=1000000000)

print(billion)

#datetime.datetime(2022, 3, 2, 1, 46, 40)

print(billion.strftime('%d/%m/%Y'))

#'02/03/2022'

How to make months not exceed 12

2 Answers2