Javascript date regex DD/MM/YYYY

Question

I know there are a lot of regex threads out there by I need a specific pattern I couldn't fin anywhere

This regex validates in a YYYY-MM-DD format

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

I need the pattern to be DD/MM/YYYY (day first since it's in spanish and only "/", "-" should not be allowed)

I searched several regex libraries and I think this one should work... but since I'm not familiar with regex I'm not sure it validates like that

(0[1-9]|[12][0-9]|3[01])[ \.-](0[1-9]|1[012])[ \.-](19|20|)\d\d

I also don't know ho to escape the slashes, I try to "see" the logic in the string but it's like trying "see" the Matrix code for me. I'm placing the regex string in a options .js

[...]  },
"date": {
                    "regex": (0[1-9]|[12][0-9]|3[01])[ \.-](0[1-9]|1[012])[ \.-](19|20|)\d\d,
                    "alertText": "Alert text AAAA-MM-DD"
                },
"other type..."[...]

So, if the regex is ok, how would I escape it? if it's not, what's the correct regex and how do I escape it? :P

Thanks a lot

Possible duplicate of [Regex to validate date format dd/mm/yyyy](http://stackoverflow.com/questions/15491894/regex-to-validate-date-format-dd-mm-yyyy) — rene, Sep 03 '16 at 19:34
Use ```(((0[1-9]|[12][0-9]|3[01])([/])(0[13578]|10|12)([/])(\d{4}))|(([0][1-9]|[12][0-9]|30)([/])(0[469]|11)([/])(\d{4}))|((0[1-9]|1[0-9]|2[0-8])([/])(02)([/])(\d{4}))|((29)(\/)(02)([/])([02468][048]00))|((29)([/])(02)([/])([13579][26]00))|((29)([/])(02)([/])([0-9][0-9][0][48]))|((29)([/])(02)([/])([0-9][0-9][2468][048]))|((29)([/])(02)([/])([0-9][0-9][13579][26])))``` to validate date in DD/MM/YYYY format, Even this will works for Leay year validation such as, this will be validate 29/02/2020 but not 29/02/2019 and also not 29/02/2100 — Harendra Chauhan, Apr 17 '20 at 16:18

score 119 · Accepted Answer · answered Mar 28 '11 at 21:42

119

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

answered Mar 28 '11 at 21:42

mVChr

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I tried flip it around but I wasn't sure where did each block started. Can this regex be edited to allow / / but not - - ? Thanks! – Juan Ignacio Mar 30 '11 at 04:32
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`^(0?[1-9]|[12][0-9]|3[01])[\/](0?[1-9]|1[012])[\/\-]\d{4}$` Regex if you just want to accept `DD/MM/YYYY` and not `DD-MM-YYYY` – lukehillonline Aug 23 '16 at 09:01
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No match on 06/20/2016 either. – Paul Sep 15 '16 at 16:31
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See @OammieR's answer for a more comprehensive regex. http://stackoverflow.com/a/14807681/246616 – mVChr Oct 07 '16 at 17:22

maerics · Answer 2 · 2013-08-02T15:19:20.163

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").

Nice idea! Only problem I found was '08/08/2013' won't pass the nonRolling validation because of the zeros. — Chris Haines, Aug 02 '13 at 10:03
No need to rebuild the string. Just `d.getDate() == m[1] && d.getMonth() == m[2] && d.getFullYear() == m[3]` — Richard Ayotte, May 24 '15 at 19:25
/^(\d{1,2})\/(\d{1,2})\/(?:\d{4}|\d{2})$/ ... I also wanted to pass dates having two or 4 digits. Nothing special simply addition at the end (?:\d{4}|\d{2}). I then also wanted to pass dates that contained hyphens or backslashes [/|-] so the FULL syntax ended looking like this... ^(\d{1,2})[/|-](\d{1,2})[/|-](?:\d{4}|\d{2})$ — Code Novice, Nov 11 '20 at 23:35

score 45 · Answer 3 · edited Feb 12 '20 at 19:30

45

Take a look from here https://www.regextester.com/?fam=114662

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]|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])|(([1][26]|[2468][048]|[3579][26])00))))$/g;

Example

edited Feb 12 '20 at 19:30

Roman Kharitonov

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answered Feb 11 '13 at 07:29

OammieR

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This works like a charm. Tested a lot of dates and its works with leap years and also for months with 30 or 31 days. – peter Apr 27 '13 at 10:06
2

Unfortunately this fails for years earlier than 1900. For instance 01/01/1899 fails. Am I missing something here? – Anjana Silva Jun 23 '15 at 08:33
5

what if I want to check this format DD/MM/YYYY instead of MM/DD/YYYY? – Ne AS Mar 15 '17 at 10:51

score 19 · Answer 4 · edited Jun 20 '20 at 09:12

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
}

There are many valid dates that does not pass your regExp test — Alex, Feb 01 '18 at 08:56
You wrote all this nice code yet none of it works when you run it. Did you check your work before posting? — Craig, Apr 20 '22 at 16:24

edid · Answer 5 · 2013-09-23T20:12:05.307

I use this function for dd/mm/yyyy format :

// (new Date()).fromString("3/9/2013") : 3 of september
// (new Date()).fromString("3/9/2013", false) : 9 of march
Date.prototype.fromString = function(str, ddmmyyyy) {
    var m = str.match(/(\d+)(-|\/)(\d+)(?:-|\/)(?:(\d+)\s+(\d+):(\d+)(?::(\d+))?(?:\.(\d+))?)?/);
    if(m[2] == "/"){
        if(ddmmyyyy === false)
            return new Date(+m[4], +m[1] - 1, +m[3], m[5] ? +m[5] : 0, m[6] ? +m[6] : 0, m[7] ? +m[7] : 0, m[8] ? +m[8] * 100 : 0);
        return new Date(+m[4], +m[3] - 1, +m[1], m[5] ? +m[5] : 0, m[6] ? +m[6] : 0, m[7] ? +m[7] : 0, m[8] ? +m[8] * 100 : 0);
    }
    return new Date(+m[1], +m[3] - 1, +m[4], m[5] ? +m[5] : 0, m[6] ? +m[6] : 0, m[7] ? +m[7] : 0, m[8] ? +m[8] * 100 : 0);
}

score 9 · Answer 6 · edited Aug 17 '15 at 14:02

9

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)

edited Aug 17 '15 at 14:02

MarmiK

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answered Aug 17 '15 at 13:20

bkaancelen

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Consider providing an explanation to your regexp – arghtype Aug 17 '15 at 13:41

score 3 · Answer 7 · answered Jul 02 '15 at 06:06

((?=\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

score 1 · Answer 8 · answered Jun 23 '15 at 14:37

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]|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))))$/

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

Thanks @OammieR for the initial idea.

score 1 · Answer 9 · answered Mar 28 '11 at 23:36

1

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.

answered Mar 28 '11 at 23:36

Naraen

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I need regex because I'm using http://www.position-absolute.com/articles/jquery-form-validator-because-form-validation-is-a-mess/ – Juan Ignacio Mar 30 '11 at 04:32
Could you elaborate since that article is huge? – jwize Apr 21 '14 at 05:54
Date.parse doesn't seem to work with UK date format (dd/mm/yyyy) on Chrome. It only works with US format (mm/dd/yyyy). – Jeff S. Mar 10 '15 at 01:52

score 0 · Answer 10 · edited May 25 '15 at 08:22

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"

score 0 · Answer 11 · answered Sep 04 '15 at 17:16

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.

score 0 · Answer 12 · answered Nov 17 '20 at 03:58

0

This validates date like dd-mm-yyyy

([0-2][0-9]|(3)[0-1])(\-)(((0)[0-9])|((1)[0-2]))(\-)([0-9][0-9][0-9][0-9])

This can use with javascript like angular reactive forms

answered Nov 17 '20 at 03:58

Visakh V A

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score 0 · Answer 13 · answered Sep 09 '22 at 15:15

It can be done like this for dd/mm/yyyy:

^(3[01]|[12][0-9]|0[1-9])/(1[0-2]|0[1-9])/[0-9]{4}$

For mm/dd/yy, mm/dd/yyyy, dd/mm/yy, and dd/mm/yyyy:

Allowing leading zeros to be omitted:

^[0-3]?[0-9]/[0-3]?[0-9]/(?:[0-9]{2})?[0-9]{2}$

Requiring leading zeros:

^[0-3][0-9]/[0-3][0-9]/(?:[0-9][0-9])?[0-9][0-9]$

For more details: https://www.oreilly.com/library/view/regular-expressions-cookbook/9781449327453/ch04s04.html

Javascript date regex DD/MM/YYYY

13 Answers13

Explanation:

Linked

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