JS replace digits with regular expressions

Question

I have an identifiers of the elements, which look like this:

form_book_1_2_3

What I want is to replace second digit in this identifier with other value. I used function 'match' with the following regexp:

 var regexp = /\d_(\d)_\d/;

But it returns me array containing

[1_2_3, 2]

So, when I try to use function 'replace' it actually replace '1_2_3' string instead of only '2'. I don't understand how to point to the second matched element in the replace function. Or maybe my regexp is wrong?

that's expected. the first entry in your matched results is always the entire string that caused the match. your actual captured values start at position [1]. — Marc B
– Marc B, Commented Jul 18, 2014 at 21:30
@MarcB ok, and how to tell "replace" function to work with value at position[1]? — Tamara
– Tamara, Commented Jul 18, 2014 at 21:32

p.s.w.g · Accepted Answer · 2014-07-18 21:37:05Z

5

The replace function will replace the entire matched string, not just a specific capture group. The easy solution is to capture the parts of the string that you don't want to modify and use them as references in the replacement pattern, for example:

"form_book_1_2_3".replace(/(\d)_\d_(\d)/, "$1_X_$2"); // "form_book_1_X_3"

For reference, the full syntax for replacement patterns is (from MDN):

$$ — Inserts a "$".

$& — Inserts the matched substring.

$` — Inserts the portion of the string that precedes the matched substring.

$' — Inserts the portion of the string that follows the matched substring.

$n or $nn — Where n or nn are decimal digits, inserts the nth parenthesized submatch string, provided the first argument was a RegExp object.

Alternatively, you could use a function to accomplish the same goal, like this:

var replacer = function(match, a, b) { return a + "_X_" + b; };
"form_book_1_2_3".replace(/(\d)_\d_(\d)/, replacer); // "form_book_1_X_3"

edited Jul 18, 2014 at 21:37

answered Jul 18, 2014 at 21:32

p.s.w.g

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2 Comments

Jongware Over a year ago

Just a footnote: I thought "shouldn't that be \1?" Looking it up proved me wrong.

Tamara Over a year ago

Thanks for providing information about $ sign. Very helpful.

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