CODEWITHSUNDEEP
javascriptregex
Gemeney
Gemeney

Reputation: 11

Match every word except if it contains a literal dot

How can I match all words except if they contain a dot, like

I want to match everything except.this and similar words with dots in them

I tried \b(?!\w+\.\w+)\w+\b but this didn't work.

No matter how I use \w+, \\., etc. the regex engine still matches the part "ignore.me" behind the dot. Is there a simple syntax for it? Just escaping the dot doesn't seem to work.

Upvotes: 0

Views: 108

Answers (1)

wp78de
wp78de

Reputation: 19000

I suggest the following pattern:

(?:^|\s)(?:(?!\.)[\w'])+(?=\s|$|[.?!](?:\s|$))

JS/Regex test: 

const regex = /(?:^|\s)(?:(?!\.)[\w'])+(?=\s|$|[.?!](?:\s|$))/g;
const str = `aaa  blabla fasdfdsa ignoremenot.
bbb igno.reme ad
It's fine?`;
let m;

while ((m = regex.exec(str)) !== null) {
// This is necessary to avoid infinite loops with zero-width matches
if (m.index === regex.lastIndex) {
    regex.lastIndex++;
}

// The result can be accessed through the `m`-variable.
m.forEach((match, groupIndex) => {
    console.log(`Found match, group ${groupIndex}: ${match.trim()}`);
});
}

There is a catch: you have to trim the matches to remove unnecessary whitespaces that can show up since we cannot use lookbehind in JavaScript's regex, like this: (?<=^|\s)(?:(?!\.)[\w'])+(?=\s|$|[.?!](?:\s|$))

Upvotes: 1

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