Python Programming: add years to current time

Question

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)

You can adapt [this answer](https://stackoverflow.com/questions/13685201/how-to-add-hours-to-current-time-in-python/13685221#13685221) to use `timedelta(years=5, days=5)` ... — Jon Clements, Aug 30 '17 at 12:39
@JonClements `class datetime.timedelta(days=0, seconds=0, microseconds=0, milliseconds=0, minutes=0, hours=0, weeks=0)` from the docs https://docs.python.org/3/library/datetime.html#datetime.timedelta. there seems to be no years argument. I think you're wrong. — void, Aug 30 '17 at 12:44
@s_vishnu ooops - confusing that with dateutil's relativedelta... — Jon Clements, Aug 30 '17 at 12:45
Possible duplicate of [Add one year in current date PYTHON](https://stackoverflow.com/questions/15741618/add-one-year-in-current-date-python) — yinnonsanders, Aug 30 '17 at 12:52
You're talking about days, but all you show is years… so `datetime.now().year + 5`…?! — deceze, Aug 30 '17 at 12:56
@Jon The point is about if OP is only interested in the year itself. — deceze, Aug 30 '17 at 13:00
@deceze sure... simple in that case.. but the *and 5 days later* not so much. — Jon Clements, Aug 30 '17 at 13:01

score 4 · Answer 1 · answered Aug 30 '17 at 12:47

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)

score 3 · Answer 2 · answered Aug 30 '17 at 12:52

3

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]>

answered Aug 30 '17 at 12:52

bruno desthuilliers

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Always forget about arrow and it's quite nice that you can do `arrow.utcnow().shift(years=5, days=5)` directly... – Jon Clements Aug 30 '17 at 12:55

Eduardo · Answer 3 · 2017-08-30T13:21:03.803

0

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.

edited Aug 30 '17 at 13:21

answered Aug 30 '17 at 12:44

Eduardo

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Never heard of leap years ? – bruno desthuilliers Aug 30 '17 at 12:47
You might also want to check your logic... This gives you 2022-07-22... – Jon Clements Aug 30 '17 at 13:14
If you're going to do it this way - make it way less complicated (and correct) and use `day_count = sum(365 + isleap(yr) for yr in range(X.year + 1, X.year + 6)) + 5` – Jon Clements Aug 30 '17 at 13:15
@Eduardo you're welcome - you only need the +5 in one place though. – Jon Clements Aug 30 '17 at 13:19
@JonClements Thanks, I overlooked it – Eduardo Aug 30 '17 at 13:21

score 0 · Answer 4 · answered Aug 31 '17 at 09:15

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

Python Programming: add years to current time

4 Answers4