Regular expression to match a word but not inside backticks

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Here is an example of different paragraphs

Upgrade is the first word in this paragraph.
In this paragraph, upgrade is the last word.
And this paragraph endsupgrade with upgrade.
But I don't want to upgradefind that word in this command `gigalixir:upgrade`.

As you can see there are 6 instances of upgrade word in above mentioned four lines. I am trying to find all the upgrade words except the one at the last (because that word is inside the command inside backticks). I also do not want to find upgrade words that are not independent.

So in the above sentence following words marked with double * should be selected:

**Upgrade** is the first word in this paragraph.
In this paragraph, **upgrade** is the last word.
And this paragraph endsupgrade with **upgrade**.
But I don't want to upgradefind that word in this command `gigalixir:upgrade`.

I have tried this simple regex:

/\bupgrade\b/gi

That selects all the independent words but I want to ignore the upgrade word inside backticks.

Note: I do not want to use lookahead or lookbehind, because I am executing this regex inside browser and any browser except chrome does not support that.

javascript regex
1 Answers

You can match strings inside backticks and skip them and only match a word upgrade in all other contexts as a whole word:

const text = 'Upgrade is the first word in this paragraph.\nIn this paragraph, upgrade is the last word.\nAnd this paragraph endsupgrade with upgrade.\nBut I don\'t want to upgradefind that word in this command `gigalixir:upgrade`.';
const regex = /(`[^`]*`)|\bupgrade\b/gi;
console.log(text.replace(regex, (x,y) => y || `**${x}**`));

The (`[^`]*`)|\bupgrade\b regex matches

  • (`[^`]*`) - Capturing group 1 (it will help analyze the match structure later): a backtick, zero or more chars other than a backtick, and a backtick
  • | - or
  • \bupgrade\b - a whole word upgrade (case insensitively due to i flag).

The .replace(regex, (x,y) => y || `**${x}**`) means that after a match is found, the match is passed to an arrow function where x is the whole match and y is the Group 1 value. If Group 1 value matches, its value is used to replace the match, else, the whole match is wrapped with double asterisks.

Alternatively, you may use a known workaround with a negative lookahead that will work only in case you have a paired amount of backticks in the string:

\bupgrade\b(?=(?:[^`]*`[^`]*`)*[^`]*$)

See the regex demo.

The (?=(?:[^`]*`[^`]*`)*[^`]*$) lookahead matches a location that is immediately followed with

  • (?:[^`]*`[^`]*`)* zero or more repetitions of any zero or more chars other than a backtick, followed with a backtick, then again any zero or more chars other than a backtick and again a backtick
  • [^`]* - any zero or more chars other than a backtick
  • $ - end of string.
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