捕获并替换多行字符串中出现的前N个模式

Question

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})/) ); 

Answer 1

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")); 

Answer 2

How about:

 const initialString = `1. One 2. Two 3. Three<br><br><br>End of List New line follows `; console.log(initialString.replace(/<br>?<br>/g, "\\n\\n")); // or do you mean: console.log(initialString.replace(/<br>?(<br>){1,}/g, "\\n\\n")); 

Answer 3

Maybe this regex can help you:

/(?:(([0-9])+.([\w]| ))*)<br><br>/g

It states that string to match must start with a number followed by . and some text and then your pattern < br > < br >. With ?: you create a non capturing group so you replace only < br >< br >.

Hope this helps.

 var str = `1. One 2. Two 3. Three<br><br><br>End of List New line follows `; console.log(str.replace(/(?:(([0-9])+.([\\w]| ))*)<br><br>/g, '\\n')); 

Answer 4

I think this should work for you:

str.replace(/<\\br><\\br>/g, '\\n');