Regular expression to validate username

Question

I'm trying to create a regular expression to validate usernames against these criteria:

Only contains alphanumeric characters, underscore and dot.
Underscore and dot can't be at the end or start of a username (e.g _username / username_ / .username / username.).
Underscore and dot can't be next to each other (e.g user_.name).
Underscore or dot can't be used multiple times in a row (e.g user__name / user..name).
Number of characters must be between 8 to 20.

This is what I've done so far; it sounds it enforces all criteria rules but the 5th rule. I don't know how to add the 5th rule to this:

 ^[a-zA-Z0-9]+([._]?[a-zA-Z0-9]+)*$

no , just one occurrence of underscore or dot is allowed each time — mohsen dorparasti, Aug 18 '12 at 12:15
yes , It is . I'm testing your Regular expression now , sounds working fine :) — mohsen dorparasti, Aug 18 '12 at 12:54
Can there be more than one dot if separated by non-dots? i.e. `"A.B.C"` (same question for underscore?) — ridgerunner, Aug 18 '12 at 13:25

Ωmega · Accepted Answer · 2020-04-30T14:18:12.657

436

^(?=.{8,20}$)(?![_.])(?!.*[_.]{2})[a-zA-Z0-9._]+(?<![_.])$
 └─────┬────┘└───┬──┘└─────┬─────┘└─────┬─────┘ └───┬───┘
       │         │         │            │           no _ or . at the end
       │         │         │            │
       │         │         │            allowed characters
       │         │         │
       │         │         no __ or _. or ._ or .. inside
       │         │
       │         no _ or . at the beginning
       │
       username is 8-20 characters long

If your browser raises an error due to lack of negative look-behind support, use the following alternative pattern:

^(?=[a-zA-Z0-9._]{8,20}$)(?!.*[_.]{2})[^_.].*[^_.]$

edited Apr 30 '20 at 14:18

answered Aug 18 '12 at 13:52

Ωmega

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Nice one. But you may want to look into using free-spacing mode and embed your comments right in there with the code. +1 – ridgerunner Aug 18 '12 at 14:23
20

@ridgerunner - I believe this way it is easy to read and port to code. – Ωmega Aug 18 '12 at 14:24
@Ωmega can we use [\w\.] instead of [a-zA-Z0-9._] – vignesh kumar Sep 03 '15 at 17:26
3

@vigneshkumar - same meaning, but `[a-zA-Z0-9._]` is easy to read and understand – Ωmega Sep 03 '15 at 21:05
2

the `<` in `(?<![_.])$` causes a parsing error for me "invalid regular expression" – Jackson Lenhart Dec 13 '16 at 16:04
3

@JacksonLenhart - What regex engine and programming language you use? Please note this question was tagged with `asp.net` tag and while the answer is valid for most of other languages and engines, not all support negative lookahead and lookbehind zero-length assertions. – Ωmega Dec 14 '16 at 12:41
Ah yes, I was naive enough to think regex would be the same for all languages. I'm using javascript (es6), no additional regex engine – Jackson Lenhart Dec 14 '16 at 15:49
3

Please note that this regular expression causes an Angular app to crash on IOS devices, as well as on the Firefox browser. – Karan Jagtiani Feb 02 '20 at 08:06
1

For those who will use this in JS: this RegEx works great in Chrome. But it raises an error in Safari, Firefox, Edge, etc. – ElRey777 Feb 12 '20 at 04:45
Firefox considers this regex to be a syntax error, and produces `SyntaxError: invalid regexp group`. – user229044 Feb 14 '20 at 15:28
wow, that is very helpful that you specified the sections. thanks so much – theProCoder Nov 25 '21 at 13:04
Thanks @Ωmega It seems that also run in Flutter for android. I did not tested on ios. – Richardd Dec 13 '21 at 00:25
First one didn't work with Safari but second one fixed it. Nice. – LacOniC Aug 20 '22 at 10:52

score 16 · Answer 2 · edited Aug 18 '12 at 13:37

16

I guess you'd have to use Lookahead expressions here. http://www.regular-expressions.info/lookaround.html

Try

^[a-zA-Z0-9](_(?!(\.|_))|\.(?!(_|\.))|[a-zA-Z0-9]){6,18}[a-zA-Z0-9]$

[a-zA-Z0-9] an alphanumeric THEN (

_(?!\.) a _ not followed by a . OR

\.(?!_) a . not followed by a _ OR

[a-zA-Z0-9] an alphanumeric ) FOR

{6,18} minimum 6 to maximum 18 times THEN

[a-zA-Z0-9] an alphanumeric

(First character is alphanum, then 6 to 18 characters, last character is alphanum, 6+2=8, 18+2=20)

edited Aug 18 '12 at 13:37

Adam Wolski

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answered Aug 18 '12 at 11:51

PhilMasteG

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Try: `^[a-zA-Z0-9](_(?!(\.|_))|\.(?!(_|\.))|[a-zA-Z0-9]){6,18}[a-zA-Z0-9]$` – Adam Wolski Aug 18 '12 at 13:27
3

Yup, this one works. But the `(_|\.)` expression can be streamlined to just `[_.]`. – ridgerunner Aug 18 '12 at 14:15
Nice, if somebody need such a RegEx for JS, this is the way. It works in all browsers unlike the most voted answer that's Chrome-only – ElRey777 Feb 12 '20 at 04:46

score 15 · Answer 3 · answered Aug 18 '12 at 13:45

As much as I love regular expressions I think there is a limit to what is readable

So I would suggest

new Regex("^[a-z._]+$", RegexOptions.IgnoreCase).IsMatch(username) &&
!username.StartsWith(".") &&
!username.StartsWith("_") &&
!username.EndsWith(".") &&
!username.EndsWith("_") &&
!username.Contains("..") &&
!username.Contains("__") &&
!username.Contains("._") &&
!username.Contains("_.");

It's longer but it won't need the maintainer to open expresso to understand.

Sure you can comment a long regex but then who ever reads it has to rely on trust.......

score 9 · Answer 4 · answered Aug 18 '12 at 12:36

9

A slight modification to Phillip's answer fixes the latest requirement

^[a-zA-Z0-9]([._](?![._])|[a-zA-Z0-9]){6,18}[a-zA-Z0-9]$

answered Aug 18 '12 at 12:36

Patrick McDonald

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This gives issue in character limit rule, with the given rule of `{6,18}` it accepts min 8 characters whereas it must be accepting 6, If I make it `{4,18}` then it will accept 6 characters whereas I have defined 4 – Ibrahim Azhar Armar Aug 22 '20 at 13:10

score 4 · Answer 5 · answered Oct 06 '18 at 12:12

private static final Scanner scan = new Scanner(System.in);

public static void main(String[] args) {
    int n = Integer.parseInt(scan.nextLine());
    while (n-- != 0) {
        String userName = scan.nextLine();
        String regularExpression = "^[[A-Z]|[a-z]][[A-Z]|[a-z]|\\d|[_]]{7,29}$";
        if (userName.matches(regularExpression)) {
            System.out.println("Valid");
        } else {
            System.out.println("Invalid");
        }
    }
}

FrEqDe · Answer 6 · 2019-11-08T19:36:09.750

3

^(?=.{4,20}$)(?:[a-zA-Z\d]+(?:(?:\.|-|_)[a-zA-Z\d])*)+$

You can test the regex here

edited Nov 08 '19 at 19:36

answered Nov 08 '19 at 17:37

FrEqDe

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This is allowing underscore at the beginning and at the end, which is not wanted. – Toto Nov 08 '19 at 17:47
2

It also matches infinite length string that contains only underscore. – Toto Nov 08 '19 at 17:50
1

@Toto, I fixed it. – FrEqDe Nov 08 '19 at 19:43
1

Works fine, but allows `-` character, I made a little adjustment to it here: ``^(?=.{4,20}$)(?:[a-zA-Z\d]+(?:[._][a-zA-Z\d])*)+$`` which fix it for me, thanks for the code @FrEqDe – Eazy Jun 21 '21 at 14:36

score 1 · Answer 7 · answered Apr 17 '17 at 15:38

1

^[a-z0-9_-]{3,15}$

^ # Start of the line

[a-z0-9_-] # Match characters and symbols in the list, a-z, 0-9, underscore, hyphen

{3,15} # Length at least 3 characters and maximum length of 15

$ # End of the line

answered Apr 17 '17 at 15:38

Mehul Jariwala

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score 1 · Answer 8 · edited Oct 11 '21 at 11:43

1

const regex = /^moe_(app|lab)[A-Za-z0-9]{3}$/;
const str = `moe_app_baa`;
let m;

if ((m = regex.exec(str)) !== null) {
    // The result can be accessed through the `m`-variable.
    m.forEach((match, groupIndex) => {
        console.log(`Found match, group ${groupIndex}: ${match}`);
    });
}

edited Oct 11 '21 at 11:43

Yu Hao

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answered Oct 11 '21 at 09:48

guisse224

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Your answer could be improved with additional supporting information. Please [edit] to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers [in the help center](/help/how-to-answer). – Community Oct 11 '21 at 09:55

bobble bubble · Answer 9 · 2022-07-02T00:38:36.580

Stumbled upon this question in another one. Seems not necessary to use that many lookarounds. Following a bit shorter pattern with use of \w which is a short for word character and commonly preset to [A-Za-z0-9_]. To get rid of the underscore, [^\W_] can be used to match [A-Za-z0-9].

^[^\W_](?!.*?[._]{2})[\w.]{6,18}[^\W_]$

^[^\W_] one letter or digit at the start
(?!.*?[._]{2}) negative lookahead to disallow __ ._ _. ..
(for reduing steps on longer inputs, a lazy quantifier is used in .*?)
[\w.]{6,18} 6 to 18 characters that can be letters, digits, _ and .
[^\W_]$ ending in another letter or digit

See this demo at regex101 (further explanation on the right side)

Note: \w depending on environment/settings can further contain unicode letters and digits.

score 0 · Answer 10 · 2012-08-20T11:02:44.357

Err sorry I generated this from my own library and it uses the syntax valid for Dart/Javascript/Java/Python, but anyway, here goes:

(?:^)(?:(?:[abcdefghijklmnopqrstuvwxyzABCDEFGHIJKlMNOPQRSTUVWXYZ0123456789]){1,1})(?!(?:(?:(?:(?:_\.){1,1}))|(?:(?:(?:__){1,1}))|(?:(?:(?:\.\.){1,1}))))(?:(?:(?:(?:[abcdefghijklmnopqrstuvwxyzABCDEFGHIJKlMNOPQRSTUVWXYZ0123456789._]){1,1})(?!(?:(?:(?:(?:_\.){1,1}))|(?:(?:(?:__){1,1}))|(?:(?:(?:\.\.){1,1}))))){6,18})(?:(?:[abcdefghijklmnopqrstuvwxyzABCDEFGHIJKlMNOPQRSTUVWXYZ0123456789]){1,1})(?:$)

My library code:

var alphaNumeric = ["a", "b", "c", "d", "e", "f", "g", "h", "i", "j", "k", "l", "m", "n", "o", "p", "q", "r", "s", "t", "u", "v", "w", "x", "y", "z", "A", "B", "C", "D", "E", "F", "G", "H", "I", "J", "K", "l", "M", "N", "O", "P", "Q", "R", "S", "T", "U", "V", "W", "X", "Y", "Z", "0", "1", "2", "3", "4", "5", "6", "7", "8", "9"];
var allValidCharacters = new List.from(alphaNumeric);
allValidCharacters.addAll([".", "_"]);

var invalidSequence = (r) => r
  .eitherString("_.")
  .orString("__")
  .orString("..");

var regex = new RegExpBuilder()
  .start()
  .exactly(1).from(alphaNumeric).notBehind(invalidSequence)
  .min(6).max(18).like((r) => r.exactly(1).from(allValidCharacters).notBehind(invalidSequence))
  .exactly(1).from(alphaNumeric)
  .end()
  .getRegExp();

My library: https://github.com/thebinarysearchtree/RegExpBuilder

ya it is hilarious :) I would probably just do a simple regular expression and then use contains, startsWith, etc... anyway. OK will fix the numbering. — , Aug 18 '12 at 12:51

score 0 · Answer 11 · answered Aug 18 '12 at 13:40

This one should do the trick:

if (Regex.IsMatch(text, @"
    # Validate username with 5 constraints.
    ^                          # Anchor to start of string.
    # 1- only contains alphanumeric characters , underscore and dot.
    # 2- underscore and dot can't be at the end or start of username,
    # 3- underscore and dot can't come next to each other.
    # 4- each time just one occurrence of underscore or dot is valid.
    (?=[A-Za-z0-9]+(?:[_.][A-Za-z0-9]+)*$)
    # 5- number of characters must be between 8 to 20.
    [A-Za-z0-9_.]{8,20}        # Apply constraint 5.
    $                          # Anchor to end of string.
    ", RegexOptions.IgnorePatternWhitespace))
{
    // Successful match
} else {
    // Match attempt failed
}

score -2 · Answer 12 · edited Oct 20 '20 at 20:43

-2

function isUserName(val){
    let regUser=/^[a-zA-Z0-9](_(?!(\.|_))|\.(?!(_|\.))|[a-zA-Z0-9]){6,18}[a-zA-Z0-9]$/;
    if(!regUser.test(val)){
      return 'Name can only use letters,numbers, minimum length is 8 characters';
    }
  }

edited Oct 20 '20 at 20:43

tjjjohnson

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answered Oct 20 '20 at 17:29

Dilusha Amarasekara

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Regular expression to validate username

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