Question 50834 in Javascript Question posted in
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How can I treat special characters as part of a whole word in a regex search? – Javascript

If I want to do a "whole word" search for a string with special characters, like say "A+", how can I write a Javascript regex that will match the whole word for cases like the following:

  1. Is the only text in the string: "A+"

  1. Is at the end of the string: "You got an A+"

  2. Is at the beginning of the string: "A+ good job!"

  3. Is in the middle: "You did A+ work here."

It should return false for:

  1. Not whole word: "A++"

  2. Not whole word": "You got an A++!"

  3. Other special characters that make it not a whole word: "A+,", "A+-", "A+0", "A+=", "A+"", "A+*", "A+/", etc.

… and many other cases

I’d love to just use b:

 /b(A+)b/g

…but that seems to interact poorly with special characters (in fact I’m not sure that regex can even match anything…).

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2

Answers


  1. 0

    You can use lookarounds for this:

    (?<=^|s)A+(?=s|$)

    Here we search for A+ surrounded by whitespace symbols or line boundaries.

    const examples = ['A+', 'You got an A+', 'A+ good job!"', 'You did A+ work here.', 'A++', 'You got an A++!', 'A+, A+-, A+0, A+=, A+", A+*, A+/'];
for (let i = 0; i < examples.length; i++) {
     console.log(examples[i], /(?<=^|s)A+(?=s|$)/.test(examples[i])) 
}

    Outputs:

    A+ true debugger eval code:1:51
You got an A+ true debugger eval code:1:51
A+ good job!" true debugger eval code:1:51
You did A+ work here. true debugger eval code:1:51
A++ false debugger eval code:1:51
You got an A++! false debugger eval code:1:51
A+, A+-, A+0, A+=, A+", A+*, A+/ false

    Demo here

    EDIT: How correctly noticed @trincot you can use negative lookarounds with non-whitespace meta sequence to avoid alterations:

    (?<!S)A+(?!S)
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  2. 0

    If you have an environment where a lookbehind assertion is not supported, you can use a capture group.

    (?:^|s)(A+)(?!S)

    Explanation

    • (?:^|s) Assert the start of the string or match a whitespace character
    • (A+) Match A and +
    • (?!S) Assert a whitespace boundary to the right

    Regex demo

    const regex = /(?:^|s)(A+)(?!S)/g;
const s = `A+
You got an A+
A+ good job!
Is in the middle: "You did A+ work here."
A++
You got an A++!
A+,
A+-
A+0
A+=
A+"
A+*
A+/`;
console.log(Array.from(s.matchAll(regex), m => m[1]))
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