Find full word by part of it using regex

Question

I have a part of the word and I should find full word in the string using regular expressions. For example, I have the following text:

If it bothers you, call it a "const identifier" instead.
It doesn't matter whether you call max a const variable or a const identififfiieer. What matters...

And the part of the word: identifi . I have to find both: identifier and identififfiieer .

I tried the following regex (javascript):

[\ ,!@#$%^&*()\.\"]*(identifi.*?)[\ ,!@#$%^&*()\d\.\"]

So it searches the part of word surrounded by punctuation characters or space. Sometime this regex works fine, but in this case it also includes quote and dot int the match. What's wrong with it? Maybe there is a better idea?

Answer 1

You can use

\bidentifi.*?\b

Which means:

• Assert the position at a word boundary
• Match the characters "identifi" literally
• Match any single character that is not not a line break
  • Between zero and unlimited times, as few times as possible, expanding as needed (lazy)
• Assert the position at a word boundary

'foo "bar identifier"'.match(/\bidentifi.*?\b/g);     // ["identifier"]
'foo identififfiieer. bar'.match(/\bidentifi.*?\b/g); // ["identififfiieer"]

Answer 2

You can use \\w*identifi\\w*

\\w stands for "word character". It always matches the ASCII characters [A-Za-z0-9_] . Notice the inclusion of the underscore and digits.

Here is a demo, showing the regex and its matches.

As a side note, your original regex actually works fine if you use the capturing group:

var body = 'If it bothers you, call it a "const identifier" instead.\nIt doesn\'t matter whether you call max a const variable or a const identififfiieer. What matters...';

var reg = /[\ ,!@#$%^&*()\.\"]*(identifi.*?)[\ ,!@#$%^&*()\d\.\"]/g;
var match;

while (match = reg.exec(body)) {
    console.log('>' + match[1] + '<');
}

This outputs:

>identifier<
>identififfiieer<

Here 'sa demo for this code.