Matching repeating groups of characters

Question

I'm trying to match groups of alphanumerics, optionally separated by the dash -.

For example,

ABC
ABC-DEF
123-DEF-ABC

but not

-
-ABC
ABC-

I've managed to do this in the backend (Java) with \w+(\-?\w+)* which works well enough, but it doesn't seem to translate to JS. I've tried many permutations with various flag settings, to no avail.

What am I doing wrong? The unit tests can be found here. Setting the sticky flag seems to pass most tests, except for the ones with a dash at the end.

Thank you

you can [**^\w+(-?\w+)*$**](https://regex101.com/r/sA4A8K/2) this — Pavneet_Singh, Jan 10 '17 at 14:16
Related http://stackoverflow.com/questions/432493/how-do-you-access-the-matched-groups-in-a-javascript-regular-expression — str, Jan 10 '17 at 14:20
@str I don't see the similarity. My problem was much simpler. Thank you nonetheless — rath, Jan 10 '17 at 14:23

Nina Scholz · Accepted Answer · 2017-01-10T15:07:15.770

3

You could use

/^\w+(-\w+)*$/g

with match for start and end of the string, for it.

var strings = ['ABC', 'ABC-DEF', '123-DEF-ABC', '-', '-ABC', 'ABC-'];

strings.forEach(function (a) {
    console.log(a.match(/^\w+(-\w+)*$/g));
    // with g                         ^
});

strings.forEach(function (a) {
    console.log(a.match(/^\w+(-\w+)*$/));
    // without g                      ^
});

.as-console-wrapper { max-height: 100% !important; top: 0; }

edited Jan 10 '17 at 15:07

answered Jan 10 '17 at 14:21

Nina Scholz

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2

When using anchors(`^`, `$`), `g` flag is useless. You can simply remove it without effect. – Tushar Jan 10 '17 at 15:03
@Tushar, please see the different result – Nina Scholz Jan 10 '17 at 15:07
If you remove `g`, then `str.match()` returns the captured items as well. You have to add `g` to ignore those, or change it to `str.match(/.../)[0]` (but watch for the null returns). – James Wilkins Jan 10 '17 at 23:52

Matching repeating groups of characters

1 Answers1