CODEWITHSUNDEEP
python
user8537998
user8537998

Reputation: 9

Python Programming: add years to current time

How do I display 5 years and 5 days later to from my current time Example :

Year : 2017 
newYear: 2022

How to do it? My current time format looks like this :

import datetime
X=datetime.datetime.now()
print ("year：%s" %x.year)

Upvotes: 2

Views: 2455

Answers (4)

JerryPlayz101
JerryPlayz101

Reputation: 117

Sorry for my Late reply, I have been extremely busy these past few days. My code will first of all add 5 years to the current year, then add five days, making sure to change it if It goes over the maximum allowed. (Which is 31 for August) But you can expand it for the other months too. This is just the concept.

    import datetime
X=datetime.datetime.now()
print ("year：%s" %X.year)
newY = X.year + 5
print("new year: %s" %newY)
newD = X.day + 5
if X.month == 1 or X.month == 3  or X.month == 5 or X.month == 7 or X.month == 8 or X.month == 10 or X.month == 11 or X.month == 12:
    # Test if the Month has 31 days
    if X.day + 5 > 31:
        op1 = X.day + 5
        op2 = op1 - 31
        new = X.month + 1
        print("month: {}".format(new))

newXD = None

Upvotes: 0

Eduardo
Eduardo

Reputation: 641

This perhaps:

from calendar import isleap
from datetime import datetime, timedelta
X=datetime.now()

day_count = sum(365 + isleap(yr) for yr in range(X.year + 1, X.year + 6)) + 5

Y = X + timedelta(days=day_count)

Note: timedelta does not accepts years directly, you have to do it using days. It is not the best method but can be done this way.

Upvotes: 0

bruno desthuilliers
bruno desthuilliers

Reputation: 77902

Or you can use arrow:

>>> import arrow
>>> ar = arrow.utcnow()
>>> ar.shift(years=5, days=5)
<Arrow [2022-09-04T12:50:26.609842+00:00]>

Upvotes: 3

Jon Clements
Jon Clements

Reputation: 142106

It's simplest to use the 3rd party dateutil and relativedelta here which conveniently takes years as a delta option and will handle leap years for you, eg:

from dateutil.relativedelta import relativedelta

dt = datetime.now() + relativedelta(years=5, days=5)
# datetime.datetime(2022, 9, 4, 13, 49, 33, 650299)

Upvotes: 4

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