Regex Retrieving data after a lookahead-CodePudding

I want to be able to retrieve !$£ from this string: (?i).*!$abc£

I have written the following Regex expression:

`(?<=\(\?i\)..)[^a-zA-Z0-9\s:]*

but for some reason it only matches !$ from the string despite if the match is seperated from the lookahead it matches perfectly.

how do i fix this?

this is javascript regex

CodePudding user response：

You can use

text = text.replace(/\(\?i\)..([^\w\s:]*)\w (\W*)/, '$1$2')
// or even
text = text.replace(/\(\?i\)..([^\w\s:]*)\w /, '$1')

See the regex demo #1 / regex demo #2

Details:

$\?i$ - a (?i) string
.. - any two chars other than line break chars
([^\w\s:]*) - Group 1 ($1): zero or more chars other than word, whitespace and : chars
\w - one or more word chars
(\W*) - Group 2: zero or more non-word chars.

The replacement is the concatenation of Group 1 and Group 2.

CodePudding user response：

You may use this replace code in javascript with a regex that matches ($i).. and grabs remaining non-whitespace string in a capture group. Then we use a replacer that removes all word characters from the captured group value.

var s = '(?i).*!$abc£aa$bb'
var r = s.replace(/\(\?i\)..(\S )/, (m, g1) => {
   return g1.replace(/\w /g, '')});
console.log(r)
//=> !$£$

CodePudding user response：

/[^!$£]/g

Replace everything but !, $, and £.

const str = `(?i).*!$abc£`;

const rgx = /[^!$£]/g;

const res = str.replaceAll(rgx, '');

console.log(res);