Regular Expression Match to test for a valid year

Question

Given a value I want to validate it to check if it is a valid year. My criteria is simple where the value should be an integer with 4 characters. I know this is not the best solution as it will not allow years before 1000 and will allow years such as 5000. This criteria is adequate for my current scenario.

What I came up with is

\d{4}$

While this works it also allows negative values.

How do I ensure that only positive integers are allowed?

fwiw I created a node.js project, [to-regex-range](https://github.com/jonschlinkert/to-regex-range) to automatically create these ranges. It's harder than it might seem if you need to generate the regex to test for a range of years. — jonschlinkert, May 28 '17 at 21:25
Why limit your validation to 4 digit years? http://longnow.org/ — Dan Temple, Feb 27 '19 at 10:53

r92 · Answer 1 · 2010-12-07T09:43:15.907

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Years from 1000 to 2999

^[12][0-9]{3}$

For 1900-2099

^(19|20)\d{2}$

edited Dec 07 '10 at 09:43

answered Dec 07 '10 at 07:18

r92

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This is a much better solution – mk_89 Oct 12 '12 at 23:41
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I's better to use non-capturing group: `^(?:19|20)\d{2}$` – Eldar Agalarov Jun 21 '15 at 12:11
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How would I validate a 4 digit year between 2011 & 2099? – mcquaim May 30 '17 at 11:37
How can I add `(19|20)\d{2}` to my regex in order to validatate a birthdate format ? This is my regex `/^[0-9]{1,2}\/(0[1-9])|(1[0-2])\/[0-9]{4}$/` I want to ensure that the year has always 4 digits and begins wither 19** or 20** – ltdev Aug 10 '17 at 14:15
Use ^(19|2[0-9])\d{2}$ for years 1900 - 2999 – MarcoZen May 11 '18 at 08:34

score 98 · Accepted Answer · answered Dec 07 '10 at 07:15

98

You need to add a start anchor ^ as:

^\d{4}$

Your regex \d{4}$ will match strings that end with 4 digits. So input like -1234 will be accepted.

By adding the start anchor you match only those strings that begin and end with 4 digits, which effectively means they must contain only 4 digits.

answered Dec 07 '10 at 07:15

codaddict

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Damn! Mine was so flawed that it would have even accepted `whateverblahblah2323`. Now I understand why a little learning is dangerous :O – Ranhiru Jude Cooray Dec 07 '10 at 07:21
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This will break in the year 10,000. – sferik Sep 03 '12 at 00:14
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@sferik: Doesn't matter. The original poster specifically stated he wanted to check for four characters. He did not state he wanted the year 10,000 to be valid input, so accepting `10000` would be a bug. – markusk Feb 01 '15 at 09:01
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@sferik: The program will be gone for centuries by then. If not, it is best to worry about 9,000 years in the future. Following the principle of YAGNI, the approach is valid. – Phil Jan 05 '18 at 18:07
@Phil not just the program, will be highly doubtful that the human race or the planet will be around at that time. And for sure robots will be doing most of the coding long before that and laughing to each other as they try to rewrite our terrible code. – blissweb Aug 22 '21 at 02:50

sferik · Answer 3 · 2013-11-16T19:27:02.600

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The "accepted" answer to this question is both incorrect and myopic.

It is incorrect in that it will match strings like 0001, which is not a valid year.

It is myopic in that it will not match any values above 9999. Have we already forgotten the lessons of Y2K? Instead, use the regular expression:

^[1-9]\d{3,}$

If you need to match years in the past, in addition to years in the future, you could use this regular expression to match any positive integer:

^[1-9]\d*$

Even if you don't expect dates from the past, you may want to use this regular expression anyway, just in case someone invents a time machine and wants to take your software back with them.

Note: This regular expression will match all years, including those before the year 1, since they are typically represented with a BC designation instead of a negative integer. Of course, this convention could change over the next few millennia, so your best option is to match any integer—positive or negative—with the following regular expression:

^-?[1-9]\d*$

edited Nov 16 '13 at 19:27

answered Sep 03 '12 at 00:34

sferik

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No, it likely won't. But someone (archeologist? historian?) might need to match years for 8000 years ago. :D And Doc Brown might need it... – Jaime Jun 24 '14 at 12:59
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i used this to limit range from 1000-9999 ^[1-9]\\d{3}$ thats because the application we have to provide data to, accept 4 digits numbers only :( – MatPag Nov 16 '15 at 11:42
Awesome answer. I'd make it the following to accept the year 0 as well: ^(-?[1-9]\d*|0)$ – Kosta Kontos Dec 15 '16 at 14:07
I would say 0001 is a valid year. And there is no way in hell this code is gonna be running in the year 2080 let alone 9999. y2k was 1960 looking forward 40 years, not looking forward 7000+ But yeah ... apart from that, some useful tidbits – blissweb Aug 22 '21 at 02:45
This will also match `1234567` so better omit the `,` -> `^[1-9]\d{3}$` – nonNumericalFloat May 03 '22 at 12:40

jonschlinkert · Answer 4 · 2015-02-01T09:51:01.740

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This works for 1900 to 2099:

/(?:(?:19|20)[0-9]{2})/

edited Feb 01 '15 at 09:51

answered Feb 01 '15 at 07:44

jonschlinkert

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looks to me like it would only go to 2099. BTW the OP is asking how to only allow positive 4 digit number. – DeanOC Feb 01 '15 at 09:07
bah, 2099 is what I meant. thanks. and now I see the positive integers part. – jonschlinkert Feb 01 '15 at 09:50
I created a project, [to-regex-range](https://github.com/jonschlinkert/to-regex-range) to automatically create these ranges – jonschlinkert Aug 17 '15 at 01:21

score 6 · Answer 5 · answered Jun 27 '13 at 09:35

6

Building on @r92 answer, for years 1970-2019:

(19[789]\d|20[01]\d)

answered Jun 27 '13 at 09:35

Renaud

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Your answer allows `197999` – avalanche1 Sep 08 '16 at 16:06
I don't think it does... The first part of the regex matches number starting with 19, then any of 7,8 or 9, followed by a SINGLE number. The regex `19[789]\d\d\d` would allow `197999 ` – Renaud Sep 09 '16 at 07:16
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`/(19[789]\d|20[01]\d)/.test(1970324) >> true` – avalanche1 Sep 12 '16 at 11:25
yes, it will match the first 4 numbers (`1970`), not the last 3. What about `(19[789]\d|20[01]\d)[^0-9]`? That would match `1970 324` but not `1970324` – Renaud Sep 13 '16 at 07:35
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I found this one useful for finding the movie year from a filename... in this case we want stuff from 1930s onwards... but we then exclude the 1024 that is added to the end for video quality – Adrian Hum Jan 28 '20 at 23:04

Dhyey · Answer 6 · 2017-06-12T12:43:22.747

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To test a year in a string which contains other words along with the year you can use the following regex: \b\d{4}\b

edited Jun 12 '17 at 12:43

answered Jun 09 '17 at 12:18

Dhyey

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This was exactly what I needed, the accepted answer doesn't seem to be valid PCRE. – Hashim Aziz Nov 28 '18 at 23:52

score 3 · Answer 7 · answered Apr 16 '18 at 08:59

In theory the 4 digit option is right. But in practice it might be better to have 1900-2099 range.

Additionally it need to be non-capturing group. Many comments and answers propose capturing grouping which is not proper IMHO. Because for matching it might work, but for extracting matches using regex it will extract 4 digit numbers and two digit (19 and 20) numbers also because of paranthesis.

This will work for exact matching using non-capturing groups:

(?:19|20)\d{2}

MarcoZen · Answer 8 · 2018-05-11T09:06:04.117

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Use;

^(19|[2-9][0-9])\d{2}$

for years 1900 - 9999.

No need to worry for 9999 and onwards - A.I. will be doing all programming by then !!! Hehehehe

You can test your regex at https://regex101.com/

Also more info about non-capturing groups ( mentioned in one the comments above ) here http://www.manifold.net/doc/radian/why_do_non-capture_groups_exist_.htm

edited May 11 '18 at 09:06

answered May 11 '18 at 08:52

MarcoZen

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score 1 · Answer 9 · answered Apr 18 '17 at 09:49

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/^\d{4}$/ This will check if a string consists of only 4 numbers. In this scenario, to input a year 989, you can give 0989 instead.

answered Apr 18 '17 at 09:49

Dayz

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score 1 · Answer 10 · answered Dec 07 '10 at 07:17

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you can go with sth like [^-]\d{4}$: you prevent the minus sign - to be before your 4 digits.
you can also use ^\d{4}$ with ^ to catch the beginning of the string. It depends on your scenario actually...

answered Dec 07 '10 at 07:17

pierroz

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score 0 · Answer 11 · answered Jan 22 '15 at 06:41

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I use this regex in Java ^(0[1-9]|1[012])[/](0[1-9]|[12][0-9]|3[01])[/](19|[2-9][0-9])[0-9]{2}$

Works from 1900 to 9999

answered Jan 22 '15 at 06:41

R2Rivs

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score 0 · Answer 12 · edited Dec 22 '14 at 14:06

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You could convert your integer into a string. As the minus sign will not match the digits, you will have no negative years.

edited Dec 22 '14 at 14:06

bluish

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answered Dec 07 '10 at 07:19

Comradin

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score 0 · Answer 13 · answered Jun 19 '18 at 15:09

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If you need to match YYYY or YYYYMMDD you can use:

^((?:(?:(?:(?:(?:[1-9]\d)(?:0[48]|[2468][048]|[13579][26])|(?:(?:[2468][048]|[13579][26])00))(?:0?2(?:29)))|(?:(?:[1-9]\d{3})(?:(?:(?:0?[13578]|1[02])(?:31))|(?:(?:0?[13-9]|1[0-2])(?:29|30))|(?:(?:0?[1-9])|(?:1[0-2]))(?:0?[1-9]|1\d|2[0-8])))))|(?:19|20)\d{2})$

answered Jun 19 '18 at 15:09

Benjamin Goodacre

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How can convert YYYYMMDD or DDMMYYYY into YYYY.MM.DD or DD.MM.YYYY. I have string like 20200315 and I want to convert to 2020.03.15 in Dart. Also incoming string might be 17042021 then I need to convert to 17.04.2020 in Dart. Thanks – NTMS Mar 04 '21 at 07:37

score 0 · Answer 14 · answered Aug 20 '19 at 11:45

0

You can also use this one.

([0-2][0-9]|3[0-1])\/([0-1][0-2])\/(19[789]\d|20[01]\d)

answered Aug 20 '19 at 11:45

Ranjeet Chouhan

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They want to match year only. Moreover your regex matches `00/00/2000` or `31/02/1999` and many more false dates. – Toto Aug 20 '19 at 12:36
Is might help you ([0-2][0-9]|3[0-1])\/([0-1][0-2])\/(19[789]\d|20[01]\d) – Ranjeet Chouhan Aug 22 '19 at 12:30
```(0[1-9]|1[0-2])``` is a much better expression for the month – itsezc Sep 26 '19 at 01:53

Leniel Maccaferri · Answer 15 · 2021-01-09T15:31:19.153

0

In my case I wanted to match a string which ends with a year (4 digits) like this for example:

Oct 2020
Nov 2020
Dec 2020
Jan 2021

It'll return true with this one:

var sheetName = 'Jan 2021';
var yearRegex = new RegExp("\b\d{4}$");
var isMonthSheet = yearRegex.test(sheetName);
Logger.log('isMonthSheet = ' + isMonthSheet);

The code above is used in Apps Script.

Here's the link to test the Regex above: https://regex101.com/r/SzYQLN/1

edited Jan 09 '21 at 15:31

answered Jan 07 '21 at 13:45

Leniel Maccaferri

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This, not only is going to match `Oct 2020` but also `Oct 20202`, `Oct20202` – Ranhiru Jude Cooray Jan 07 '21 at 21:04
@RanhiruJudeCooray improved the regex. – Leniel Maccaferri Jan 09 '21 at 15:31

score 0 · Answer 16 · answered Sep 09 '21 at 18:18

0

You can try the following to capture valid year from a string:

.*(19\d{2}|20\d{2}).*

answered Sep 09 '21 at 18:18

greenie

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This allows arbitrary text before and after the year, which is not what the question is looking for. – joanis Sep 09 '21 at 19:28

score -1 · Answer 17 · answered May 02 '22 at 20:50

-1

Works from 1950 to 2099 and value is an integer with 4 characters

^(?=.*?(19[56789]|20\d{2}).*)\d{4}$

answered May 02 '22 at 20:50

Алина Пчельникова

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Regular Expression Match to test for a valid year

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