Vfrdtyky

You could take the regex that validates YYYY/MM/DD and flip it around to get what you need for DD/MM/YYYY:

/^(0?[1-9]|[12][0-9]|3[01])[/-](0?[1-9]|1[012])[/-]d{4}$/

BTW - this regex validates for either DD/MM/YYYY or DD-MM-YYYY

P.S. This will allow dates such as 31/02/4899

A regex is good for matching the general format but I think you should move parsing to the Date class, e.g.:

function parseDate(str) {

  var m = str.match(/^(d{1,2})/(d{1,2})/(d{4})$/);

  return (m) ? new Date(m[3], m[2]-1, m[1]) : null;

}

Now you can use this function to check for valid dates; however, if you need to actually validate without rolling (e.g. "31/2/2010" doesn't automatically roll to "3/3/2010") then you've got another problem.

[Edit] If you also want to validate without rolling then you could add a check to compare against the original string to make sure it is the same date:

function parseDate(str) {

  var m = str.match(/^(d{1,2})/(d{1,2})/(d{4})$/)

    , d = (m) ? new Date(m[3], m[2]-1, m[1]) : null

    , nonRolling = (d&&(str==[d.getDate(),d.getMonth()+1,d.getFullYear()].join('/')));

  return (nonRolling) ? d : null;

}

[Edit2] If you want to match against zero-padded dates (e.g. "08/08/2013") then you could do something like this:

function parseDate(str) {

  function pad(x){return (((''+x).length==2) ? '' : '0') + x; }

  var m = str.match(/^(d{1,2})/(d{1,2})/(d{4})$/)

    , d = (m) ? new Date(m[3], m[2]-1, m[1]) : null

    , matchesPadded = (d&&(str==[pad(d.getDate()),pad(d.getMonth()+1),d.getFullYear()].join('/')))

    , matchesNonPadded = (d&&(str==[d.getDate(),d.getMonth()+1,d.getFullYear()].join('/')));

  return (matchesPadded || matchesNonPadded) ? d : null;

}

However, it will still fail for inconsistently padded dates (e.g. "8/08/2013").

Take a look from here http://forums.asp.net/t/1410702.aspx/1

Use this following Regular Expression Details, This will support leap year also.

var reg = /^(((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]|1d|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))))$/g;

Example

Scape slashes is simply use before / and it will be escaped. (/=> /).

Otherwise you're regex DD/MM/YYYY could be next:

/^[0-9]{2}[/]{1}[0-9]{2}[/]{1}[0-9]{4}$/g

Explanation:

• 
  [0-9]: Just Numbers


• 
  {2} or {4}: Length 2 or 4. You could do {2,4} as well to length between two numbers (2 and 4 in this case)


• 
  [/]: Character /
  


• 
  g : Global -- Or m: Multiline (Optional, see your requirements)


• 
  $: Anchor to end of string. (Optional, see your requirements)


• 
  ^: Start of string. (Optional, see your requirements)

An example of use:

var regex = /^[0-9]{2}[/][0-9]{2}[/][0-9]{4}$/g;



var dates = ["2009-10-09", "2009.10.09", "2009/10/09", "200910-09", "1990/10/09", 

    "2016/0/09", "2017/10/09", "2016/09/09", "20/09/2016", "21/09/2016", "22/09/2016",

    "23/09/2016", "19/09/2016", "18/09/2016", "25/09/2016", "21/09/2018"];



//Iterate array

dates.forEach(

    function(date){

        console.log(date + " matches with regex?");

      console.log(regex.test(date));

    });

Of course you can use as boolean:

if(regex.test(date)){

     //do something

}

Try using this..

[0-9]{2}[/][0-9]{2}[/][0-9]{4}$

this should work with this pattern DD/DD/DDDD where D is any digit (0-9)

If you are in Javascript already, couldn't you just use Date.Parse() to validate a date instead of using regEx.

RegEx for date is actually unwieldy and hard to get right especially with leap years and all.

((?=d{4})d{4}|(?=[a-zA-Z]{3})[a-zA-Z]{3}|d{2})((?=/)/|-)((?=[0-9]{2})[0-9]{2}|(?=[0-9]{1,2})[0-9]{1,2}|[a-zA-Z]{3})((?=/)/|-)((?=[0-9]{4})[0-9]{4}|(?=[0-9]{2})[0-9]{2}|[a-zA-Z]{3})

Regex Compile on it

2012/22/Jan

2012/22/12 

2012/22/12

2012/22/12

2012/22/12

2012/22/12

2012/22/12

2012-Dec-22

2012-12-22

23/12/2012

23/12/2012

Dec-22-2012

12-2-2012

23-12-2012

23-12-2012

For people who needs to validate years earlier than year 1900, following should do the trick. Actually this is same as the above answer given by [@OammieR][1] BUT with years including 1800 - 1899.

/^(((0[1-9]|[12]d|3[01])/(0[13578]|1[02])/((19|[2-9]d)d{2}))|((0[1-9]|[12]d|3[01])/(0[13578]|1[02])/((18|[2-9]d)d{2}))|((0[1-9]|[12]d|30)/(0[13456789]|1[012])/((19|[2-9]d)d{2}))|((0[1-9]|[12]d|30)/(0[13456789]|1[012])/((18|[2-9]d)d{2}))|((0[1-9]|1d|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))))$/

Hope this helps someone who needs to validate years earlier than 1900, such as 01/01/1855, etc.

Thanks @OammieR for the initial idea.

Do the following change to the jquery.validationengine-en.js file and update the dd/mm/yyyy inline validation by including leap year:

"date": {

    // Check if date is valid by leap year

    "func": function (field) {

    //var pattern = new RegExp(/^(d{4})[/-.](0?[1-9]|1[012])[/-.](0?[1-9]|[12][0-9]|3[01])$/);

    var pattern = new RegExp(/^(0?[1-9]|[12][0-9]|3[01])[/-.](0?[1-9]|1[012])[/-.](d{4})$/);

    var match = pattern.exec(field.val());

    if (match == null)

    return false;



    //var year = match[1];

    //var month = match[2]*1;

    //var day = match[3]*1;

    var year = match[3];

    var month = match[2]*1;

    var day = match[1]*1;

    var date = new Date(year, month - 1, day); // because months starts from 0.



    return (date.getFullYear() == year && date.getMonth() == (month - 1) && date.getDate() == day);

},

"alertText": "* Invalid date, must be in DD-MM-YYYY format"

I build this regular to check month 30/31 and let february to 29.

new RegExp(/^((0[1-9]|[12][0-9]|3[01])(/)(0[13578]|1[02]))|((0[1-9]|[12][0-9])(/)(02))|((0[1-9]|[12][0-9]|3[0])(/)(0[469]|11))(/)d{4}$/)

I think, it's more simple and more flexible and enough full.

Perhaps first part can be contract but I Don't find properly.