Why does my regex in JavaScript return undefined?

Question

I'm trying to find all the matches for 'test' in my string:

const search = "test";
const regexString = "(?:[^ ]+ ){0,3}" + "test" + "(?: [^ ]+){0,3}";
const re = new RegExp(regexString, "gi");
const matches = [];
const fullText = "my test string with a lot of tests that should match the test regex";
let match = re.exec(fullText);
while (match != undefined) {
    matches.push(match[1]);
    match = re.exec(fullText);
}
console.log(matches);

I'm getting the following:

[ undefined, undefined, undefined ]

Why isn't my search working?

Trying to return an array containing the 3 words preceding and proceeding 'test' — Mary
– Mary, Commented Apr 16, 2018 at 22:27
Just to clarify, if exec(..) doesn't found a match it will return null instead of undefined. — Rodrigo Ferreira
– Rodrigo Ferreira, Commented Apr 16, 2018 at 22:35
I'm not an expert of regular expressions, but if you substitute matches.push(match[1]) with matches.push(match[0]) you obtain: ["my test string with a", "lot of test", "should match the test regex"] — Francesco Blue
– Francesco Blue, Commented Apr 16, 2018 at 22:47

Pointy · Accepted Answer · 2018-04-16 22:28:05Z

4

Your code expects the result of the match to include stuff captured in capturing groups in the regular expression. However, your regular expression contains only non-capturing groups. The (?: ) grouping explicitly does not capture the matched substring.

You want plain ( ) groupings.

answered Apr 16, 2018 at 22:28

Pointy

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2 Comments

Mary Over a year ago

Thanks. When I change to: const regexString = "([^ ]+ ){0,3}" + "test" + "( [^ ]+){0,3}"; I get: [ 'my ', 'of ', 'the ' ]. Is there a way to capture a few words before and after the 'test' search search? Sorry for beginner question regex confuses me.

Pointy Over a year ago

@Mary right - your code expects there to be something in match[1], but if there are no real capturing groups match will always be an array with just one value. Your code currently only looks at match[1], but the regular expression (fixed) captures two groups, so you need to also look at match[2].

revo · Accepted Answer · 2018-04-16 22:34:32Z

1

You should enclose your non-capturing groups (?:...) in a capturing group (...) since you are invoking a capturing group (match[1]). :

"((?:\\S+ ){0,3})" + search + "((?: \\S+){0,3})"

Trying to return an array containing the 3 words preceding and proceeding 'test'

Then you need to push both captured groups not one:

matches.push([match[1], search, match[2]]);
// `match[1]` refers to first capturing group
// `match[2]` refers to second CG
// `search` contains search word

JS code:

const search = "test";
const regexString = "((?:\\S+ ){0,3})" + search + "((?: \\S+){0,3})";
const re = new RegExp(regexString, "gi");
const matches = [];
const fullText = "my test string with a lot of tests that should match the test regex";
while ((match = re.exec(fullText)) != null) {
    matches.push([match[1], search, match[2]]);
}
console.log(matches);

answered Apr 16, 2018 at 22:34

revo

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3 Comments

revo Over a year ago

This, however, doesn't output correct data on overlapping matches. A workaround would be using a lookahead construct.

Rodrigo Ferreira Over a year ago

Is worth mentioning that regex doesn't support repeated capturing groups so the best approach is to capture the three words (from before and after) together in a bigger capturing group.

revo Over a year ago

@RodrigoFerreira It's not the regex, it's the flavor. i.e. .NET supports accessing to quantified capturing groups, each one individually.

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