Regular expression for only characters a-z, A-Z

Question

I don't know how to create a regular expression in JavaScript or jQuery.

I want to create a regular expression that will check if a string contains only characters between a-z and A-Z with any arrangement.

EDIT

When I tried to make regex

/^[a-zA-Z\s]+$/

to accept white spaces as well. It is not working. What could be the mistake?

I am testing my regular expression at JavaScript RegExp Example: Online Regular Expression Tester.

On your new regex, it should work. It is working here, at least. — NullUserException, Aug 20 '10 at 15:12
See: http://rubular.com/r/kl8wklKBxy (I know it's ruby, but I tested it on JS as well) — NullUserException, Aug 20 '10 at 15:13
Your new regex will not work on that site if you included the `/`s, as those delimit the start and end of the expression. — Ian Henry, Aug 20 '10 at 15:21

score 79 · Accepted Answer · edited Jun 20 '20 at 09:12

79

/^[a-zA-Z]*$/

Change the * to + if you don't want to allow empty matches.

References:

Character classes ([...]), Anchors (^ and $), Repetition (+, *)

The / are just delimiters, it denotes the start and the end of the regex. One use of this is now you can use modifiers on it.

edited Jun 20 '20 at 09:12

Community

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answered Aug 20 '10 at 14:56

NullUserException

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why ^ and $ has been used. What is its purpose – Shantanu Gupta Aug 20 '10 at 14:59
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@Shantanu: Those are used to match the beginning and the end of the string. Without them any string containing letters would be matched. Now only strings containing ONLY letters will be matched. – Jakob Aug 20 '10 at 15:00
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@Shantanu: ^ is begin of string, $ is end of string. It's used here to make sure the complete string does contain characters. If you would do `/[a-zA-Z]*/` if would also match the string "1234 fo!" (at "fo"). – DarkDust Aug 20 '10 at 15:01

Hooray Im Helping · Answer 2 · 2015-04-19T15:08:08.847

Piggybacking on what the other answers say, since you don't know how to do them at all, here's an example of how you might do it in JavaScript:

var charactersOnly = "This contains only characters";
var nonCharacters = "This has _@#*($()*@#$(*@%^_(#@!$ non-characters";

if (charactersOnly.search(/[^a-zA-Z]+/) === -1) {
  alert("Only characters");
}

if (nonCharacters.search(/[^a-zA-Z]+/)) {
  alert("There are non characters.");
}

The / starting and ending the regular expression signify that it's a regular expression. The search function takes both strings and regexes, so the / are necessary to specify a regex.

From the MDN Docs, the function returns -1 if there is no match.

Also note: that this works for only a-z, A-Z. If there are spaces, it will fail.

Ollie Edwards · Answer 3 · 2010-08-20T15:10:52.457

13

/^[a-zA-Z]+$/

Off the top of my head.

Edit:

Or if you don't like the weird looking literal syntax you can do it like this

new RegExp("^[a-zA-Z]+$");

edited Aug 20 '10 at 15:10

answered Aug 20 '10 at 14:55

Ollie Edwards

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I missed the ^ and $, I'll edit my answer to make it correct but props to @NullUserException for getting it first – Ollie Edwards Aug 20 '10 at 15:10

score 2 · Answer 4 · edited Oct 16 '18 at 09:21

2

With POSIX Bracket Expressions (not supported by Javascript) it can be done this way:

/[:alpha:]+/

Any alpha character A to Z or a to z.

or

/^[[:alpha:]]+$/s

to match strictly with spaces.

edited Oct 16 '18 at 09:21

Waruyama

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answered Aug 20 '10 at 15:00

RobertPitt

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But... shouldn't this be /[[:alpha:]]+/ ? The regular expression [a-zA-Z] will return incorrect values for some character encodings (e.g. EBCDIC, where there is a gap between i and j, and another between r and s (which includes the ~ character), and between I and J and between R and S. This one works everywhere. – James McLeod Aug 20 '10 at 15:04
`[[:alpha:]]` is correct, but lots of languages will just drop a warning and DWYM. or it might just be untested. – muhmuhten Aug 20 '10 at 15:13
@James, arguably, if the question is taken literally, `[a-zA-Z]` will give correct results even for EBCDIC, and `[[:alpha:]]` will return incorrect results for EBCDIC. Because the question did not ask for only letters. It asked for a-z, A-Z. :-/ – LarsH Aug 20 '10 at 21:16
@LarsH - that's a very (in fact annoyingly) good point, but I think I'll stick with my interpretation. Who really wants to include ~, }, and a bunch of unprintables with a-zA-Z? – James McLeod Aug 20 '10 at 21:36

score 1 · Answer 5 · answered Feb 15 '21 at 16:42

This /[^a-z]/g solves the problem.

function pangram(str) {
    let regExp = /[^a-z]/g;
    let letters = str.toLowerCase().replace(regExp, '');
    document.getElementById('letters').innerHTML = letters;
}
pangram('GHV 2@# %hfr efg uor7 489(*&^% knt lhtkjj ngnm!@#$%^&*()_');

<h4 id="letters"></h4>

Regular expression for only characters a-z, A-Z

5 Answers5

References:

Linked