-
tnurnberg@mysql.com authored
if input year for date_add() / date_sub() with INTERVAL is low enough for calc_daynr() to possibly return incorrect results (calc_daynr() has no information on whether the year is low because it was a two-digit year ('77) or because it was a really low four-digit year (0077) and will indiscriminately try to turn the value into a four-digit year by adding 1900 or 2000 respectively), the functions will now throw NULL.
d1311e1a