8

I'm looking for a solution to split a string at every nth linebreak. Lets say i have one string that has six lines

"One\nTwo\nThree\nFour\nFive\nSix\n"

So splitting at 3rd line break would give me something like

"One\nTwo\nThree\n" and "Four\nFive\nSix\n"

I've found solutions to do it at nth character, but i can't be definite of at what character length the nth break would occur. I hope my question is clear. Thanks.

4
  • Don't try to split, try to match at least 3 lines. Commented Sep 29, 2017 at 17:49
  • @CasimiretHippolyte Not quite sure how to do that, i've found patterns that match multiple lines, having trouble finding patterns that match every n number of lines. Commented Sep 29, 2017 at 17:52
  • 1
    @HaiderAli In this case, How do you want your output to be when your input is One\n\n\n\n\nTwo\n\nThree\n\nFour\n\nFive\n\n\n\nSix\n? Commented Sep 29, 2017 at 18:03
  • @Gurman That will not occur, the string is programatically prepared from an array. I'd rather split a string rather paginate an array ;p Commented Sep 29, 2017 at 18:12

2 Answers 2

4

Instead of using the String.prototype.split, it's easier to use the String.prototype.match method:

"One\nTwo\nThree\nFour\nFive\nSix\n".match(/(?=[\s\S])(?:.*\n?){1,3}/g);

demo

pattern details:

(?=[\s\S]) # ensure there's at least one character (avoid a last empty match)

(?:.*\n?)  # a line (note that the newline is optional to allow the last line)

{1,3} # greedy quantifier between 1 and 3
      # (useful if the number of lines isn't a multiple of 3)

Other way with Array.prototype.reduce:

"One\nTwo\nThree\nFour\nFive\nSix\n".split(/^/m).reduce((a, c, i) => {
    i%3  ?  a[a.length - 1] += c  :  a.push(c);
    return a;
}, []);
Sign up to request clarification or add additional context in comments.

Comments

2

Straight-forward:

(?:.+\n?){3}

See a demo on regex101.com.


Broken down, this says:

(?:  # open non-capturing group
.+   # the whole line
\n?  # a newline character, eventually but greedy
){3} # repeat the group three times

Comments

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.