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How can we check the date in ddmmyyyy format (including leap year validation) using regular expression? What would be the regular expression...?

nhahtdh
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Gangadhar Jannu
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    Did you try something already? – Juru Oct 14 '14 at 08:27
  • **Can you do this without regular expression?** It would be a mess to maintain. – nhahtdh Oct 14 '14 at 08:28
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    Have you considered the possibility, that regex may not be the right solution to that problem? – Danubian Sailor Oct 14 '14 at 08:29
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    Have a look at: http://stackoverflow.com/a/3873172/372239 – Toto Oct 14 '14 at 08:33
  • I'm trying this in angularJS, I've to write another directive for this validation – Gangadhar Jannu Oct 14 '14 at 08:36
  • @M42 I've already searched over google about validation but I want validation for date in ddmmyyyy format. Can you show me the duplicated one I haven't found my answer in the link provided by you – Gangadhar Jannu Oct 14 '14 at 08:41
  • The answer I gave does exactly ddmmyyyy, but @M42's link is a very similar approach and it shouldn't be too hard for anyone to adapt it and remove the separator. You are sounding very "give me ze code" and "do my work for me" when insisting in multiple comments that you only want something that works for exactly ddmmyyyy. – asontu Oct 14 '14 at 08:44
  • @funkwurm I know the stackoverflow rules man however I've tried similar approaches by removing seperators (-, ., /) but I was unable to find the correct one. Anyway thanks for the answer. – Gangadhar Jannu Oct 14 '14 at 08:51

2 Answers2

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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))$

Regular expression visualization

Debuggex Demo

asontu
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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)$"
nhahtdh
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FaheemFayaz
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