RegExp.prototype[Symbol.match]()

Try it

class RegExp1 extends RegExp {
  [Symbol.match](str) {
    const result = RegExp.prototype[Symbol.match].call(this, str);
    if (result) {
      return "VALID";
    }
    return "INVALID";
  }
}

console.log("2012-07-02".match(new RegExp1("(\\d+)-(\\d+)-(\\d+)")));
// Expected output: "VALID"

Syntax

regexp[Symbol.match](str)

Parameters

str

A String that is a target of the match.

Return value

An Array whose contents depend on the presence or absence of the global (g) flag, or null if no matches are found.

• If the g flag is used, all results matching the complete regular expression will be returned, but capturing groups are not included.
• If the g flag is not used, only the first complete match and its related capturing groups are returned. In this case, match() will return the same result as RegExp.prototype.exec() (an array with some extra properties).

Description

This method exists for customizing match behavior within RegExp subclasses. It is called internally in String.prototype.match(). For example, the following two examples return same result.

"abc".match(/a/);

/a/[Symbol.match]("abc");

If the regex is global (with the g flag), its lastIndex is first set to 0, so matching always starts from the beginning of the string, and the regex's exec() method is repeatedly called until exec() returns null. If the current match is an empty string, the lastIndex would still be advanced — if the regex is Unicode-aware, it would advance by one Unicode code point; otherwise, it advances by one UTF-16 code unit.

console.log("😄".match(/(?:)/g)); // [ '', '', '' ]
console.log("😄".match(/(?:)/gu)); // [ '', '' ]

If the regex is not global, exec() would only be called once and its result becomes the return value of [Symbol.match]().

The exec() method automatically resets lastIndex to 0 when the last match fails, so for global regexes with lastIndex starting at 0, [Symbol.match]() generally produces no side-effects. However, when the regex is sticky but not global, exec() is only called once and therefore does not reset lastIndex if the match was successful. In this case, each call to match() may return a different result.

const re = /[abc]/y;
for (let i = 0; i < 5; i++) {
  console.log("abc".match(re), re.lastIndex);
}
// [ 'a' ] 1
// [ 'b' ] 2
// [ 'c' ] 3
// null 0
// [ 'a' ] 1

When the regex is sticky and global, it would still perform sticky matches — i.e., it would fail to match any occurrences beyond the lastIndex.

console.log("ab-c".match(/[abc]/gy)); // [ 'a', 'b' ]

In addition, the [Symbol.match] property is used to check whether an object is a regular expression.

Examples

Direct call

This method can be used in almost the same way as String.prototype.match(), except the different this and the different arguments order.

const re = /\d+/g;
const str = "2016-01-02";
const result = re[Symbol.match](str);
console.log(result); // ["2016", "01", "02"]

Using [Symbol.match]() in subclasses

Subclasses of RegExp can override the [Symbol.match]() method to modify the default behavior.

class MyRegExp extends RegExp {
  [Symbol.match](str) {
    const result = RegExp.prototype[Symbol.match].call(this, str);
    if (!result) return null;
    return {
      group(n) {
        return result[n];
      },
    };
  }
}

const re = new MyRegExp("(\\d+)-(\\d+)-(\\d+)");
const str = "2016-01-02";
const result = str.match(re); // String.prototype.match calls re[Symbol.match]().
console.log(result.group(1)); // 2016
console.log(result.group(2)); // 01
console.log(result.group(3)); // 02

Specifications

Browser compatibility

See also

• Polyfill of RegExp.prototype[Symbol.match] in core-js
• String.prototype.match()
• RegExp.prototype[Symbol.matchAll]()
• RegExp.prototype[Symbol.replace]()
• RegExp.prototype[Symbol.search]()
• RegExp.prototype[Symbol.split]()
• RegExp.prototype.exec()
• RegExp.prototype.test()
• Symbol.match