How to handle Python datetime.replace overflow case?

Question

I've got case when I should add an hour to datetime object and handle situation when such addition leads to day addition. E.g. it is 23:59 1st September, I add 1 hour and got 0:59 2nd September. Standard Python replace function doesn't handle such case and throws ValueError:

>>> import datetime
>>> dt = datetime.datetime.now()
>>> dt
datetime.datetime(2019, 12, 2, 19, 20, 29, 850677)
>>> dt.replace(hour=dt.hour+10)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
ValueError: hour must be in 0..23

Of course I could write custom function which handles such situation, but I'd like to avoid it.

Could anybody help?

Jefferson Houp · Accepted Answer · 2019-12-02 14:26:35Z

3

There we have to use timedelta:

from datetime import datetime, timedelta

dt = datetime.now()
dt2 = dt + timedelta(hours=10)

answered Dec 2, 2019 at 14:26

Jefferson Houp

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Alex Hall · Accepted Answer · 2019-12-02 14:26:12Z

3

dt += datetime.timedelta(hours=10)

answered Dec 2, 2019 at 14:26

Alex Hall

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