en.javascript.info/9-regular-expressions/10-regexp-backreferences/article.md

Backreferences in pattern: \n and \k

Capturing groups can be accessed not only in the result or in the replacement string, but also in the pattern itself.

Backreference by number: \n

A group can be referenced in the pattern using \n, where n is the group number.

To make things clear let's consider a task.

We need to find a quoted string: either a single-quoted subject:'...' or a double-quoted subject:"..." -- both variants need to match.

How to look for them?

We can put two kinds of quotes in the pattern: pattern:['"](.*?)['"], but it would find strings with mixed quotes, like match:"...' and match:'...". That would lead to incorrect matches when one quote appears inside other ones, like the string subject:"She's the one!":

let str = `He said: "She's the one!".`;

let reg = /['"](.*?)['"]/g;

// The result is not what we expect
alert( str.match(reg) ); // "She'

As we can see, the pattern found an opening quote match:", then the text is consumed lazily till the other quote match:', that closes the match.

To make sure that the pattern looks for the closing quote exactly the same as the opening one, we can make a groups of it and use the backreference.

Here's the correct code:

let str = `He said: "She's the one!".`;

*!*
let reg = /(['"])(.*?)\1/g;
*/!*

alert( str.match(reg) ); // "She's the one!"

Now it works! The regular expression engine finds the first quote pattern:(['"]) and remembers the content of pattern:(...), that's the first capturing group.

Further in the pattern pattern:\1 means "find the same text as in the first group", exactly the same quote in our case.

Please note:

• To reference a group inside a replacement string -- we use $1, while in the pattern -- a backslash \1.
• If we use ?: in the group, then we can't reference it. Groups that are excluded from capturing (?:...) are not remembered by the engine.

Backreference by name: \k<name>

For named groups, we can backreference by \k<name>.

The same example with the named group:

let str = `He said: "She's the one!".`;

*!*
let reg = /(?<quote>['"])(.*?)\k<quote>/g;
*/!*

alert( str.match(reg) ); // "She's the one!"