Regex for special characters working wrong

Question

I'm trying to create regex for checking if password got special characters. https://www.owasp.org/index.php/Password_special_characters It looks like this

new RegExp('[!@#\$%\^\&*\)\(+=._\'",/<>?[\\\]`{|}~:;-\s]', 'g');

Unfortunately is also catching bare words: reg.test('word') it returns true. Whats wrong with my regex?

It should return true only if there is one or more of special characters in string. — user9010407, Jul 06 '18 at 09:13
reg.test('word') should return false; reg.test('word!') should return true — user9010407, Jul 06 '18 at 09:14
Because `[;-s]` matches a lot of letters. ***Double escape** backslashes in a string literal to define a *regex escape*. Better, use a **regex literal**, `/regex_pattern_here/flags_here`. — Wiktor Stribiżew, Jul 06 '18 at 09:17

Alex · Answer 1 · 2018-07-06T09:24:51.380

0

You can use negative/inverse logic and test against any character that is not a number or a letter.

Using [^A-Za-z0-9] where caret (^) matches everything except A-Za-z0-9.

const regex = /([^A-Za-z0-9]|[.\p{L}])/gm;

console.log('word: ' + regex.test('word'));
console.log('word!: ' + regex.test('word!'));
console.log('!£2word!: ' + regex.test('!£2word!'));
console.log('!ą,ć,ó!"£wordąćó: ' + regex.test('!ą,ć,ó!"£wordąćó'));

edited Jul 06 '18 at 09:24

answered Jul 06 '18 at 09:17

Alex

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It's almost great , thanks a lot but how to ignore Diacritic characters like ą,ć,ó ? – user9010407 Jul 06 '18 at 09:21
@user9010407 edited. – Alex Jul 06 '18 at 09:24

score 0 · Answer 2 · answered Jul 06 '18 at 09:17

You didn't escape properly. RegExp object could receive a regular expression as a string but escaping matters: it needs double backslashes.

For now [;-\s] is equal to [;-s] which includes 57 characters:

[_;?\[\]@\\`\^<->aAbBcCdDeEfFgGhHiIjJkKlLmMnNoOpPqQrRsS-Z]

However, it should be [;\-\\s].

Regex for special characters working wrong

2 Answers2