[removed] negative lookbehind equivalent?

Question

Is there a way to achieve the equivalent of a negative lookbehind in javascript regular expressions? I need to match a string that does not start with a specific set of cha

Answer 1

Since 2018, Lookbehind Assertions are part of the ECMAScript language specification.

// positive lookbehind
(?<=...)
// negative lookbehind
(?<!...)

---

Answer pre-2018

As Javascript supports negative lookahead, one way to do it is:

1. reverse the input string

2. match with a reversed regex

3. reverse and reformat the matches

---

const reverse = s => s.split('').reverse().join('');

const test = (stringToTests, reversedRegexp) => stringToTests
  .map(reverse)
  .forEach((s,i) => {
    const match = reversedRegexp.test(s);
    console.log(stringToTests[i], match, 'token:', match ? reverse(reversedRegexp.exec(s)[0]) : 'Ø');
  });

Example 1:

Following @andrew-ensley's question:

test(['jim', 'm', 'jam'], /m(?!([abcdefg]))/)

Outputs:

jim true token: m
m true token: m
jam false token: Ø

Example 2:

Following @neaumusic comment (match max-height but not line-height, the token being height):

test(['max-height', 'line-height'], /thgieh(?!(-enil))/)

Outputs:

max-height true token: height
line-height false token: Ø

Answer 2

/(?![abcdefg])[^abcdefg]m/gi yes this is a trick.

Answer 3

Let's suppose you want to find all int not preceded by unsigned:

With support for negative look-behind:

(?<!unsigned )int

Without support for negative look-behind:

((?!unsigned ).{9}|^.{0,8})int

Basically idea is to grab n preceding characters and exclude match with negative look-ahead, but also match the cases where there's no preceeding n characters. (where n is length of look-behind).

So the regex in question:

(?<!([abcdefg]))m

would translate to:

((?!([abcdefg])).|^)m

You might need to play with capturing groups to find exact spot of the string that interests you or you want to replace specific part with something else.

Answer 4

Mijoja's strategy works for your specific case but not in general:

js>newString = "Fall ball bill balll llama".replace(/(ba)?ll/g,
   function($0,$1){ return $1?$0:"[match]";});
Fa[match] ball bi[match] balll [match]ama

Here's an example where the goal is to match a double-l but not if it is preceded by "ba". Note the word "balll" -- true lookbehind should have suppressed the first 2 l's but matched the 2nd pair. But by matching the first 2 l's and then ignoring that match as a false positive, the regexp engine proceeds from the end of that match, and ignores any characters within the false positive.

Answer 5

Use

newString = string.replace(/([abcdefg])?m/, function($0,$1){ return $1?$0:'m';});

Answer 6

Lookbehind Assertions got accepted into the ECMAScript specification in 2018.

Positive lookbehind usage:

console.log(
  "$9.99  €8.47".match(/(?<=\$)\d+(\.\d*)?/) // Matches "9.99"
);

Negative lookbehind usage:

console.log(
  "$9.99  €8.47".match(/(?<!\$)\d+(?:\.\d*)/) // Matches "8.47"
);

Platform support:

• ✔️ V8
  • ✔️ Google Chrome 62.0
  • ✔️ Microsoft Edge 79.0
  • ✔️ Node.js 6.0 behind a flag and 9.0 without a flag
  • ✔️ Deno (all versions)
• ✔️ SpiderMonkey
  • ✔️ Mozilla Firefox 78.0