Lookahead assertion: (?=...), (?!...)

pattern

A pattern consisting of anything you may use in a regex literal, including a disjunction.

A regular expression generally matches from left to right. This is why lookahead and lookbehind assertions are called as such — lookahead asserts what's on the right, and lookbehind asserts what's on the left.

In order for a (?=pattern) assertion to succeed, the pattern must match the text after the current position, but the current position is not changed. The (?!pattern) form negates the assertion — it succeeds if the pattern does not match at the current position.

The pattern can contain capturing groups. See the capturing groups page for more information on the behavior in this case.

Unlike other regular expression operators, there's no backtracking into a lookahead — this behavior is inherited from Perl. This only matters when the pattern contains capturing groups and the pattern following the lookahead contains backreferences to those captures. For example:

/(?=(a+))a*b\1/.exec("baabac"); // ['aba', 'a']
// Not ['aaba', 'a']

The matching of the pattern above happens as follows:

1. The lookahead (a+) succeeds before the first "a" in "baabac", and "aa" is captured because the quantifier is greedy.
2. a*b matches the "b" in "baabac".
3. \1 does not match the following string, because that requires 2 "a"s, but only 1 is available. So the matcher backtracks, but it doesn't go into the lookahead, so the capturing group cannot be reduced to 1 "a", and the entire match fails at this point.
4. exec() re-attempts matching at the next position — before the second "a". This time, the lookahead matches "a", and a*b matches "b". The backreference \1 matches the captured "a", and the match succeeds.

If the regex is able to backtrack into the lookahead and revise the choice made in there, then the match would succeed at step 3 by (a+) matching the first "a" (instead of the first two "a"s) and a*b matching "ab", without even re-attempting the next input position.

Negative lookaheads can contain capturing groups as well, but backreferences only make sense within the pattern, because if matching continues, pattern would necessarily be unmatched (otherwise the assertion fails). This means outside of the pattern, backreferences to those capturing groups in negative lookaheads always succeed. For example:

/(.*?)a(?!(a+)b\2c)\2(.*)/.exec("baaabaac"); // ['baaabaac', 'ba', undefined, 'abaac']

The matching of the pattern above happens as follows:

1. The (.*?) pattern is non-greedy, so it starts by matching nothing. However, the next character is a, which fails to match "b" in the input.
2. The (.*?) pattern matches "b" so that the a in the pattern matches the first "a" in "baaabaac".
3. At this position, the lookahead succeeds to match, because if (a+) matches "aa", then (a+)b\2c matches "aabaac". This causes the assertion to fail, so the matcher backtracks.
4. The (.*?) pattern matches the "ba" so that the a in the pattern matches the second "a" in "baaabaac".
5. At this position, the lookahead fails to match, because the remaining input does not follow the pattern "any number of "a"s, a "b", the same number of "a"s, a c". This causes the assertion to succeed.
6. However, because nothing was matched within the assertion, the \2 backreference has no value, so it matches the empty string. This causes the rest of the input to be consumed by the (.*) at the end.

Normally, assertions cannot be quantified. However, in Unicode-unaware mode, lookahead assertions can be quantified. This is a deprecated syntax for web compatibility, and you should not rely on it.

/(?=a)?b/.test("b"); // true; the lookahead is matched 0 time

Sometimes it's useful to validate that the matched string is followed by something without returning that as the result. The following example matches a string that is followed by a comma/period, but the punctuation is not included in the result:

function getFirstSubsentence(str) {
  return /^.*?(?=[,.])/.exec(str)?.[0];
}

getFirstSubsentence("Hello, world!"); // "Hello"
getFirstSubsentence("Thank you."); // "Thank you"

A similar effect can be achieved by capturing the submatch you are interested in.

Using lookahead, you can match a string multiple times with different patterns, which allows you to express complex relationships like subtraction (is X but not Y) and intersection (is both X and Y).

The following example matches any identifier that's not a reserved word (only showing three reserved words here for brevity; more reserved words can be added to this disjunction). The [$_\p{ID_Start}][$\u200c\u200d\p{ID_Continue}]* syntax describes exactly the set of identifier strings in the language spec; you can read more about identifiers in lexical grammar and the \p escape in Unicode character class escape.

function isValidIdentifierName(str) {
  const re =
    /^(?!(?:break|case|catch)$)[$_\p{ID_Start}][$\u200c\u200d\p{ID_Continue}]*$/u;
  return re.test(str);
}

isValidIdentifierName("break"); // false
isValidIdentifierName("foo"); // true
isValidIdentifierName("cases"); // true

The following example matches a string that's both ASCII and can be used as an identifier part:

function isASCIIIDPart(char) {
  return /^(?=\p{ASCII}$)\p{ID_Start}$/u.test(char);
}

isASCIIIDPart("a"); // true
isASCIIIDPart("α"); // false
isASCIIIDPart(":"); // false

If you are doing intersection and subtraction with finitely many characters, you may want to use the character set intersection syntax enabled with the v flag.

• Assertions
• Regular expressions reference
• Input boundary assertion: ^, $
• Word boundary assertion: \b, \B
• Lookbehind assertion: (?<=...), (?<!...)
• Capturing group: (...)