Python Regex Date -

i'm trying create regex looks dates in format mm-dd-yyyy, , came far:

dateregex = re.compile(r'''             (0[1-9]|1[0-2])                  # month              -                                  (10|20|[0-2][1-9]|3[01])         # day: not [0-2][0-9]|3[01] avoid 00 matching              -             ((198[0-9]|20(0[0-9]|1[0-6])'''  # year: matches 1980 - 2016             , re.verbose) 

is there easier way allows me create range of numbers? , wanted create 1 allows legal dates (for example, june shouldn't have 31 days), easiest way match months different days, like:

((01|03|05|07|08|10|12)-(31 day regex pattern)-(year regex) # 31-day months | (04|06|09|11)-(30 day regex pattern)-(year regex)           # 30-day months | 02-(regex depending on leap year))                          # 28 or 29 days 

not sure how february besides putting leaps years , 29 days together, , remaining years 28 days.

i agree jonrsharpe way combine regex datetime. used simple regex going match could date in format, try parse them datetime.

import re import datetime  def yield_valid_dates(datestr):     match in re.finditer(r"\d{1,2}-\d{1,2}-\d{4}", datestr):         try:             date = datetime.datetime.strptime(match.group(0), "%m-%d-%y")             yield date             # or can yield match.group(0) if want             # yield date string found 05-04-1999         except valueerror:             # date couldn't parsed datetime... invalid date             pass   teststr = """05-04-1999 here filler text in between 2 dates 4-5-2016 invalid  date 32-2-2016 here invalid date, there no 32d day of month 6-32-2016. can not  include leading zeros 4-2-2016 , still detected"""  date in yield_valid_dates(teststr):     print(date) 

this prints 3 valid dates:

1999-05-04 00:00:00 2016-04-05 00:00:00 2016-04-02 00:00:00