[3.14] gh-70647: Better promote how to safely parse yearless dates in datetime. (GH-116179) (#143037)

gh-70647: Better promote how to safely parse yearless dates in datetime. (GH-116179)

* gh-70647: Better promote how to safely parse yearless dates in datetime.

Every four years people encounter this because it just isn't obvious.
This moves the footnote up to a note with a code example.

We'd love to change the default year value for datetime but doing
that could have other consequences for existing code.  This documented
workaround *always* works.

* doctest code within note is bad, dedent.

* Update to match the error message.

* remove no longer referenced footnote

* ignore the warning in the doctest

* use Petr's suggestion for the docs to hide the warning processing

* cover date.strptime (3.14) as well
(cherry picked from commit b8d3fddba6)

Co-authored-by: Gregory P. Smith <68491+gpshead@users.noreply.github.com>
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@ -2634,9 +2634,42 @@ Broadly speaking, ``d.strftime(fmt)`` acts like the :mod:`time` module's
``time.strftime(fmt, d.timetuple())`` although not all objects support a
:meth:`~date.timetuple` method.
For the :meth:`.datetime.strptime` class method, the default value is
``1900-01-01T00:00:00.000``: any components not specified in the format string
will be pulled from the default value. [#]_
For the :meth:`.datetime.strptime` and :meth:`.date.strptime` class methods,
the default value is ``1900-01-01T00:00:00.000``: any components not specified
in the format string will be pulled from the default value.
.. note::
When used to parse partial dates lacking a year, :meth:`.datetime.strptime`
and :meth:`.date.strptime` will raise when encountering February 29 because
the default year of 1900 is *not* a leap year. Always add a default leap
year to partial date strings before parsing.
.. testsetup::
# doctest seems to turn the warning into an error which makes it
# show up and require matching and prevents the actual interesting
# exception from being raised.
# Manually apply the catch_warnings context manager
import warnings
catch_warnings = warnings.catch_warnings()
catch_warnings.__enter__()
warnings.simplefilter("ignore")
.. testcleanup::
catch_warnings.__exit__()
.. doctest::
>>> from datetime import datetime
>>> value = "2/29"
>>> datetime.strptime(value, "%m/%d")
Traceback (most recent call last):
...
ValueError: day 29 must be in range 1..28 for month 2 in year 1900
>>> datetime.strptime(f"1904 {value}", "%Y %m/%d")
datetime.datetime(1904, 2, 29, 0, 0)
Using ``datetime.strptime(date_string, format)`` is equivalent to::
@ -2767,7 +2800,7 @@ Notes:
include a year in the format. If the value you need to parse lacks a year,
append an explicit dummy leap year. Otherwise your code will raise an
exception when it encounters leap day because the default year used by the
parser is not a leap year. Users run into this bug every four years...
parser (1900) is not a leap year. Users run into that bug every leap year.
.. doctest::
@ -2794,5 +2827,3 @@ Notes:
.. [#] See R. H. van Gent's `guide to the mathematics of the ISO 8601 calendar
<https://web.archive.org/web/20220531051136/https://webspace.science.uu.nl/~gent0113/calendar/isocalendar.htm>`_
for a good explanation.
.. [#] Passing ``datetime.strptime('Feb 29', '%b %d')`` will fail since 1900 is not a leap year.