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How to add date and years(integer) to get a date
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How to add date and years(integer) to get a date
#1
I am trying to add years in integer form in date to get final output as date. Could anyone help with syntax.
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#2
show us what have you tried, sample input and expected output

probably https://docs.python.org/3/library/dateti...te.replace or https://docs.python.org/3/library/dateti...me.replace
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#3
What have you tried? Show us your code and where you are stuck. Have you tried timedelta(days=years*365) but got stuck on leap years?
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#4
(Sep-02-2020, 07:22 PM)NG0824 Wrote: I am trying to add years in integer form in date to get final output as date. Could anyone help with syntax.

In [1]: import datetime

In [2]: now = datetime.datetime.now()

In [3]: print(now)
2020-09-03 10:01:22.062401

In [4]: # set a year
   ...: past = now.replace(year=1900)

In [5]: print(past)
1900-09-03 10:01:22.062401

In [6]: # adding years to current year
   ...: future = now.replace(now.year + 10) # 10 years in future

In [7]: leap_year = datetime.datetime(2020, 2, 29) # impossible date for non-leap years

In [8]: print(leap_year)
2020-02-29 00:00:00

In [9]: leap_year.replace(year=2019)  # booom
---------------------------------------------------------------------------
ValueError                                Traceback (most recent call last)
<ipython-input-9-164fec2883be> in <module>
----> 1 leap_year.replace(year=2019)  # booom

ValueError: day is out of range for month
If you wonder, this comes from IPython. A better REPL for Python.
Working with dates is not always straightforward. Sometimes it ends into impossible dates.

To catch this leap_error:

import datetime

one_day = datetime.timedelta(days=1)
leap_year = datetime.datetime(2020, 2, 29)
new_year = 2019

try:
    new_dt = leap_year.replace(year=new_year)
except ValueError:
    new_dt = (leap_year + one_day).replace(year=new_year)
Important stuff to read:
I hope this pushes you in the right direction.
Just keep in mind, that the replace method does not change the object itself, it returns a new modified datetime object.
My code examples are always for Python >=3.6.0
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#5
Thanks everyone, I was able to solve this error by firt installing datedelta and then used datedelat.datedelta(years=fieldname())
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