Capture and replace first N occurrence of a pattern in a multiline string

priyeshdkr Published at Dev

24

priyeshdkr

I have following string from which i expect to replace the first two occurrences of <br> with a \n character.

<br><br><br>. Do not replace <br> here.
1. One
2. Two
3. Three<br><br><br>End of List. Replace first two <br> with \n
New line follows.
<br><br><br>. Do not replace <br> here.

I did write my regex here. I am very new to regex and i am sure this is not an optimized solution. After some tries i was able to select the <br><br> as a capturing group. I want this 3rd capturing group to be my selected match so i can easly replace it with \n. Can someone help me with this?

My expected output is:

<br><br><br>. Do not replace <br> here.
1. One
2. Two
3. Three\n<br>End of List. Replace first two <br> with \n
New line follows.
<br><br><br>. Do not replace <br> here.

var str = `1. One
2. Two
3. Three<br><br><br>End of List
New line follows
`

console.log(
  str.match(/[\n\r].*(\d\.\s+)(?!.*[\n\r](\d\.\s+)).*((<br\s*\/?>){2})/)
);

Joven28

Try this.

let str = `<br><br><br>. Do not replace <br> here.
1. One
2. Two
3. Three<br><br><br>End of List. Replace first two <br> with \n
New line follows.
<br><br><br>. Do not replace <br> here.`;
console.log(str.replace(/(?<=[0-9]+\..*?)(<br>){2}/g, "\n"));

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edited at2020-11-30

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