Regex substitute every instance of the phrase CONTAINED within brackets

I am looking to replace all instances of text within brackets, including the brackets themselves, in a given string. The specific text to be replaced will be stored in a variable in Javascript. Using a simple regex in a replace method won't work due to the inclusion of brackets.

For example, if we want to replace "[test] [teste] test [hello]" with the value stored in a variable as "hi", the expected output would be "hi hi test [hello]".

Answer №1

"[apple] [banana] orange".replace(/\[.*?\]/g, 'hi')

To escape the brackets, use "\" and include the g flag

Update: The i flag has been removed and the w has been changed to . in order to handle all content inside brackets

Answer №2

Not completely certain about what you're searching for, but using .match will save the matches in an array and .replace will carry out the replacement.

const regex = /\[.*?\]/g;

var mutable = "[test] [teste] test";
const matches = mutable.match(regex); // Storing all matches in an array
mutable = mutable.replace(regex, 'dude'); // Replacing the matches

console.log(mutable);
console.log(matches);

Answer №3

After experimenting, I discovered a method to replace my variable with specific characters. For example:

var sample = "[sample]", Then I changed the brackets to "\[sample\]", and followed it up with:
var regex = new RegExp(sample+"+","gm")

Finally, I applied this regex in the JavaScript replace function.

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