Regular expression for ddmmyyyy date including validation for leap year [duplicate]

Answer 1

Regex for dd/MM/yyyy even for leap year:

pattern="(^(((0[1-9]|[12][0-8])[\/](0[1-9]|1[012]))|((29|30|31)[\/](0[13578]|1[02]))|((29|30)[\/](0[4,6,9]|11)))[\/](19|[2-9][0-9])\d\d$)|(^29[\/]02[\/](19|[2-9][0-9])(00|04|08|12|16|20|24|28|32|36|40|44|48|52|56|60|64|68|72|76|80|84|88|92|96)$)"

Or simple

pattern= "^(0?[1-9]|[12][0-9]|3[01])/(0?[1-9]|1[012])/((19|20)\\d\\d)$"

Answer 2

Dude, you asked the question I've been working on for a couple of weeks. I invite those commenting to give a date that breaks this. Now note that this works for the years 1000-9999, is Proleptic Gregorian and assumes that we won't change how leap-years work until the year 9999 ;)

^(?:(?:(?:0[1-9]|1\d|2[0-8])(?:0[1-9]|1[0-2])|(?:29|30)(?:0[13-9]|1[0-2])|31(?:0[13578]|1[02]))[1-9]\d{3}|2902(?:[1-9]\d(?:0[48]|[2468][048]|[13579][26])|(?:[2468][048]|[13579][26])00))$

Debuggex Demo