How to replace RegExp capture groups?
javascript, regex
Solution
The problem is caused by the two layers of interpretation that the backslashes undergo. First you have JavaScript's string literal syntax, which uses backslashes to escape characters. When you have a string like
'(\'|\"|\_|\[)(A)(\'|\"|\_|\])'
in your code, the actual contents are:
('|"|_|[)(A)('|"|_|])
This is then parsed by the RegExp constructor, which treats `[)(A)('|"|_|]` as a single character class equivalent to `[A'"_()|]`.
To preserve a backslash so the regexp engine can see it, you have to put two backslashes in your string.
You can also use character classes to simplify your code a bit:
var regex = new RegExp('([\'"_[])A([\'"_\\]])', 'g')
var string = ' [A] _A_ "A" \'A\' A BAB "B" '
console.log(string.replace(regex, '$1X$2'));
Problem
In a string, I need to replace all occurences of `[A]`, `"A"`, `'A'` and `_A_` with `[X]`, `"X"`, `'X'` and `_X_`. I tried this answer but it's giving a weird result (code below). I use `new RegExp` because it allows interpolating a variable -- `A` and `X` are just an example. ``` // CAPTURE GROUPS: |------1-----|2|-----3------| var regexp = new RegExp('(\'|\"|\_|\[)(A)(\'|\"|\_|\])', 'g') var string = ' [A] _A_ "A" \'A\' A BAB "B" ' string.replace(regex, '$1X$3') // ORIGINAL: [A] _A_ "A" 'A' A BAB "B" // EXPECTED: [X] _X_ "X" 'X' A BAB "B" // ACTUAL: [X] XXX XXX XXX X BXB XBX ```