javascript regex to warn invalid username

Question

these are the conditions for a valid username

must contain a-z or A-Z
can contain 0-9 anywhere
must not allow any other characters

so a0 and 0a are valid user names just like username and UserName. i need a regex for this.

the issue optionally allowing 0-9

UPDATE the problem with regex like /\w+/i is it allows username like 000 but in my case it must contain an alphabet and not any special characters

@RenzoCC this allows `09` as username but it must contain atleast one alphabet — Haseeb A, Oct 05 '17 at 13:48
@ctwheels i searched before asking. I asked because i feel this question is unique. can you provide a link — Haseeb A, Oct 05 '17 at 13:58
@HaseebEqx My apologies, you can use `^(?=.*[a-zA-Z])[a-zA-Z0-9]+$` and here's a link to an existing question with same conditions: https://stackoverflow.com/questions/1051989/regex-for-alphanumeric-but-at-least-one-character — ctwheels, Oct 05 '17 at 14:22
`\w` may not be the best to use here since it allows for `_` and may allow for foreign characters in some languages. Also `[\w]` can be replaced with `\w` as the `[]` is redundant here — ctwheels, Oct 05 '17 at 14:32
Possible duplicate of [regex for alphanumeric, but at least one character](https://stackoverflow.com/questions/1051989/regex-for-alphanumeric-but-at-least-one-character) — glennsl, Oct 05 '17 at 14:35

Thomas Ayoub · Answer 1 · 2017-10-05T14:40:01.903

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You don't know how to compose lookahead ? just use

var isValid = /[a-zA-Z]/.test(username) && /^[a-zA-Z\d]$/.test(username);
              ^^^^^^^^^ at last one char
                                           ^^^^^^^^^^^^^only char & digits

You want to learn lookahead ? Just use:

var isValid = /^(?=.*[a-z])[a-z\d]+$/gmi.test(username);

edited Oct 05 '17 at 14:40

answered Oct 05 '17 at 13:50

Thomas Ayoub

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Andrzej Smyk · Answer 2 · 2017-10-05T14:24:00.367

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Maybe something like this:

/^[a-zA-z0-9]*[a-zA-z][a-zA-z0-9]*$/

edited Oct 05 '17 at 14:24

answered Oct 05 '17 at 14:01

Andrzej Smyk

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it allows `user+` – Haseeb A Oct 05 '17 at 14:05

kami4ka · Answer 3 · 2017-10-05T14:35:10.577

0

Simple solution:

/^(\d*[a-zA-Z]+\d*)+$/

edited Oct 05 '17 at 14:35

answered Oct 05 '17 at 14:00

kami4ka

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`[\d]` is the same as `\d`, no need for square brackets `[]`. Also there's an extra `(` at the beginning, so this is invalid – ctwheels Oct 05 '17 at 14:34

score 0 · Accepted Answer · answered Oct 05 '17 at 14:19

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the working regex is /^([0-9]+[a-zA-Z]|[a-zA-Z]+[0-9]|[a-zA-Z])+$/.

it does match 0a a0 aaa

but will not match 000 or +a or a+

please read the question completely, before putting down votes

answered Oct 05 '17 at 14:19

Haseeb A

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I have edited my answer and posted regexp that works and is a bit simpler – Andrzej Smyk Oct 05 '17 at 14:26
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Although this does work, the regex I put in my comment `^(?=.*[a-zA-Z])[a-zA-Z0-9]+$` will perform with an even number or fewer steps in all cases. The one that @AndrzejSmyk posted will use fewer steps in all cases as well – ctwheels Oct 05 '17 at 14:29

javascript regex to warn invalid username

4 Answers4