Regular Expression for Date Format : dd-mm-yyyy

Question

I want to check the date which must be in the format dd-mm-yyyy using a regular expression, and it also must check the leap year dates.

I am using RegularExpressionValidator for checking the date.

score 2 · Answer 1 · answered Oct 29 '13 at 18:18

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try this. It works for me!

ValidationExpression="(^((((0[1-9])|([1-2][0-9])|(3[0-1]))|([1-9]))-(((0[1-9])|(1[0-2]))|([1-9]))-(([0-9]{2})|(((19)|([2]([0]{1})))([0-9]{2}))))$)"

answered Oct 29 '13 at 18:18

fmaga

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Only answer the worked for me, tried the others, didn't work, ty! – Alex Apr 14 '15 at 00:54

score 1 · Accepted Answer · answered Feb 01 '13 at 06:54

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Try this regular expression-

^(((((0[1-9])|(1\d)|(2[0-8]))-((0[1-9])|(1[0-2])))|((31-((0[13578])|(1[02])))|((29|30)-((0[1,3-9])|(1[0-2])))))-((20[0-9][0-9]))|(29-02-20(([02468][048])|([13579][26]))))$

Got it from Here

answered Feb 01 '13 at 06:54

Microsoft DN

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score 1 · Answer 3 · edited May 29 '13 at 18:54

This regex also handles leap year:

^(((0[1-9]|[12]\d|3[01])/(0[13578]|1[02])/((19|[2-9]\d)\d{2}))|((0[1-9]|[12]\d|30)/(0[13456789]|1[012])/((19|[2-9]\d)\d{2}))|((0[1-9]|1\d|2[0-8])/02/((19|[2-9]\d)\d{2}))|(29/02/((1[6-9]|[2-9]\d)(0[48]|[2468][048]|[13579][26])|((16|[2468][048]|[3579][26])00))))$

Matches
[29/02/2000], [30/04/2003], [01/01/2003]

Non-Matches
[29/02/2001], [30-04-2003], [1/1/1899]

You can also check this link out: http://www.codeproject.com/KB/aspnet/LengthValidation.aspx

score 0 · Answer 4 · edited May 23 '17 at 11:45

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You can javascript to check leap year for more info

isLeap = new Date(year, 1, 29).getMonth() == 1

Regular Expression

^(?:^(?:(?:(?:(?:(?:0?[13578]|1[02])/31)|(?:(?:0?[13-9]|1[0-2])/(?:29|30)))/(?:1[6-9]|[2-9]\d)\d{2})|(?:0?2/29/(?:(?:(?:1[6-9]|[2-9]\d)(?:0[48]|[2468][048]|[13579][26])|(?:(?:16|[2468][048]|[3579][26])00))))|(?:(?:0?[1-9])|(?:1[0-2]))/(?:0?[1-9]|1\d|2[0-8])/(?:(?:1[6-9]|[2-9]\d)\d{2}))$)$

These allow but do not require a leading zero in single-digit months/days. If you don't want that, replace all instances of 0? with 0.

edited May 23 '17 at 11:45

Community

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answered Feb 01 '13 at 06:40

Amol

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Parsing dates with regular expressions is possible, but frustrating. It's hard to get right, the expression is difficult for non-regex wizards to understand (which means it's difficult to prove that the thing is correct), and it is slow compared to other options. – Amol Feb 01 '13 at 06:48
see [this](http://stackoverflow.com/questions/4290433/regular-expression-for-aspregularexpressionvalidator-with-format-mmddyy-leap-y) if you dont have any option – Amol Feb 01 '13 at 06:50

score 0 · Answer 5 · answered May 29 '13 at 19:32

You could use a CustomValidator and have the client-side validation be simple and on the server-side use a DateTime.TryParse to get a definitive validation. Although I suspect you don't need your code to work all the way to the year 9999 (no, I couldn't immediately see if the supplied regexes work that far into the future).

score 0 · Answer 6 · answered Jan 11 '17 at 18:05

from Microsoft DN but modified to work with years both 20xx and 19xx to used as DOB

^(((((0[1-9])|(1\d)|(2[0-8]))/((0[1-9])|(1[0-2])))|((31-((0[13578])|(1[02])))|((29|30)-((0[1,3-9])|(1[0-2])))))/((((20[0-9][0-9]))|(29-02-20(([02468][048])|([13579][26]))))|(((19[0-9][0-9]))|(29-02-19(([02468][048])|([13579][26]))))))$

score -1 · Answer 7 · edited Jan 12 '19 at 11:34

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for dd/MM/yyyy formate

(((0|1)[0-9]|2[0-9]|3[0-1])\/(0[1-9]|1[0-2])\/((19|20)\d\d))$

edited Jan 12 '19 at 11:34

Zoe

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answered Dec 31 '18 at 14:07

Chandrakant Dewangan

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How come anything before 1900 and after 2099 is an invalid date? – Imantas Dec 31 '18 at 14:24
implement your own logic change the number in place of number you see and check its working or not – Chandrakant Dewangan Jan 04 '19 at 02:07

Regular Expression for Date Format : dd-mm-yyyy

7 Answers7