Regex to check for at least 3 characters?

Question

I have this regex to allow for only alphanumeric characters.

How can I check that the string at least contains 3 alphabet characters as well.

My current regex,

if(!/^[a-zA-Z0-9]+$/.test(val))

I want to enforce the string to make sure there is at least 3 consecutive alphabet characters as well so;

111 // false
aaa1 // true
11a // false
bbc // true
1a1aa // false

@ka that would match at least three _characters_ not at least three _letters_ in any string. — Boris the Spider, Mar 11 '13 at 14:28
{3,} means that the string must match a minimum of 3 characters. If you add a digit after the comma, that also sets a max limit. — Jenny D, Mar 11 '13 at 14:28

score 99 · Answer 1 · edited Apr 10 '20 at 21:35

99

+ means "1 or more occurrences."

{3} means "3 occurrences."

{3,} means "3 or more occurrences."

+ can also be written as {1,}.

* can also be written as {0,}.

edited Apr 10 '20 at 21:35

Sergio Carneiro

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answered Mar 11 '13 at 14:28

Andy Lester

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I don't think this answers the question, it answers a comment to the question which itself does not answer the question. – Boris the Spider Mar 11 '13 at 14:30
3

True - but still VERY helpful to understand the logic of one of the previous answer(s). – jeffkee Oct 30 '18 at 22:17
A small addition: `?` can be also written as `{0,1}` – Albin Jul 18 '23 at 22:09

Andrew Cheong · Accepted Answer · 2013-03-11T14:37:34.580

To enforce three alphabet characters anywhere,

/(.*[a-z]){3}/i

should be sufficient.

Edit. Ah, you'ved edited your question to say the three alphabet characters must be consecutive. I also see that you may want to enforce that all characters should match one of your "accepted" characters. Then, a lookahead may be the cleanest solution:

/^(?.*[a-z]{3})[a-z0-9]+$/i

Note that I am using the case-insensitive modifier /i in order to avoid having to write a-zA-Z.

Alternative. You can read more about lookaround assertions here. But it may be a little bit over your head at this stage. Here's an alternative that you may find easier to break down in terms of what you already know:

/^([a-z0-9]*[a-z]){3}[a-z0-9]*$/i

For those wishing to use this in a `grep` or `git grep` context, you'll need the `-E` option (a.k.a. `--extended-regexp`) for the `{n}` quantifier to be understood. — Quintin Willison, May 14 '19 at 11:56

Marchev · Answer 3 · 2013-03-11T14:40:01.840

10

This should do the work:

^([0-9]*[a-zA-Z]){3,}[0-9]*$

It checks for at least 3 "Zero-or-more numerics + 1 Alpha" sequences + Zero-or-more numerics.

edited Mar 11 '13 at 14:40

answered Mar 11 '13 at 14:33

Marchev

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score 0 · Answer 4 · answered Mar 11 '13 at 14:34

0

You want to match zero or more digits then 3 consecutive letters then any other number of digits?

/\d*(?:[a-zA-Z]){3,}\d*/

answered Mar 11 '13 at 14:34

Boris the Spider

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score 0 · Answer 5 · answered May 06 '22 at 11:45

0

This is vanilla JS you guys can use. My problem is solved using this.

const str = "abcdggfhf";
const pattern = "fhf";

if(pattern.length>2) {
console.log(str.search(pattern));
}

answered May 06 '22 at 11:45

Afridi

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Regex to check for at least 3 characters?

5 Answers5

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