How to remove all line breaks from a string

I have a text in a textarea and I read it out using the .value attribute.

Now I would like to remove all linebreaks (the character that is produced when you press Enter) from my text now using .replace with a regular expression, but how do I indicate a linebreak in a regex?

If that is not possible, is there another way?

This is probably a FAQ. Anyhow, line breaks (better: newlines) can be one of Carriage Return (CR, \r, on older Macs), Line Feed (LF, \n, on Unices incl. Linux) or CR followed by LF (\r\n, on WinDOS). (Contrary to another answer, this has nothing to do with character encoding.)

Therefore, the most efficient RegExp literal to match all variants is


If you want to match all newlines in a string, use a global match,


respectively. Then proceed with the replace method as suggested in several other answers. (Probably you do not want to remove the newlines, but replace them with other whitespace, for example the space character, so that words remain intact.)

How you'd find a line break varies between operating system encodings. Windows would be \r\n, but Linux just uses \n and Apple uses \r.

I found this in JavaScript line breaks:

someText = someText.replace(/(\r\n|\n|\r)/gm, "");

That should remove all kinds of line breaks.

var str = " \n this is a string \n \n \n"


String.trim() removes whitespace from the beginning and end of strings... including newlines.

const myString = "   \n \n\n Hey! \n I'm a string!!!         \n\n";
const trimmedString = myString.trim();

// outputs: "Hey! \n I'm a string!!!"

Here's an example fiddle:

NOTE! it only trims the beginning and end of the string, not line breaks or whitespace in the middle of the string.

You can use \n in a regex for newlines, and \r for carriage returns.

var str2 = str.replace(/\n|\r/g, "");

Different operating systems use different line endings, with varying mixtures of \n and \r. This regex will replace them all.

If you want to remove all control characters, including CR and LF, you can use this:

myString.replace(/[^\x20-\x7E]/gmi, "")

It will remove all non-printable characters. This are all characters NOT within the ASCII HEX space 0x20-0x7E. Feel free to modify the HEX range as needed.

The simplest solution would be:

let str = '\t\n\r this  \n \t   \r  is \r a   \n test \t  \r \n';
str.replace(/\s+/g, ' ').trim();
console.log(str); // logs: "this is a test"

.replace() with /\s+/g regexp is changing all groups of white-spaces characters to a single space in the whole string then we .trim() the result to remove all exceeding white-spaces before and after the text.

Are considered as white-spaces characters:
[ \f\n\r\t\v?\u00a0\u1680?\u2000?-\u200a\u2028\u2029\u202f\u205f\u3000\ufeff]

var str = "bar\r\nbaz\nfoo";

str.replace(/[\r\n]/g, '');

>> "barbazfoo"

To remove new line chars use this:

yourString.replace(/\r?\n?/g, '')

Then you can trim your string to remove leading and trailing spaces:


The answer provided by PointedEars is everything most of us need. But by following Mathias Bynens's answer, I went on a Wikipedia trip and found this:

The following is a drop-in function that implements everything the above Wiki page considers "new line" at the time of this answer.

If something doesn't fit your case, just remove it. Also, if you're looking for performance this might not be it, but for a quick tool that does the job in any case, this should be useful.

// replaces all "new line" characters contained in `someString` with the given `replacementString`
const replaceNewLineChars = ((someString, replacementString = ``) => { // defaults to just removing
  const LF = `\u{000a}`; // Line Feed (\n)
  const VT = `\u{000b}`; // Vertical Tab
  const FF = `\u{000c}`; // Form Feed
  const CR = `\u{000d}`; // Carriage Return (\r)
  const CRLF = `${CR}${LF}`; // (\r\n)
  const NEL = `\u{0085}`; // Next Line
  const LS = `\u{2028}`; // Line Separator
  const PS = `\u{2029}`; // Paragraph Separator
  const lineTerminators = [LF, VT, FF, CR, CRLF, NEL, LS, PS]; // all Unicode `lineTerminators`
  let finalString = someString.normalize(`NFD`); // better safe than sorry? Or is it?
  for (let lineTerminator of lineTerminators) {
    if (finalString.includes(lineTerminator)) { // check if the string contains the current `lineTerminator`
      let regex = new RegExp(lineTerminator.normalize(`NFD`), `gu`); // create the `regex` for the current `lineTerminator`
      finalString = finalString.replace(regex, replacementString); // perform the replacement
  return finalString.normalize(`NFC`); // return the `finalString` (without any Unicode `lineTerminators`)

A linebreak in regex is \n, so your script would be

var test = 'this\nis\na\ntest\nwith\newlines';
console.log(test.replace(/\n/g, ' '));

I am adding my answer, it is just an addon to the above, as for me I tried all the /n options and it didn't work, I saw my text is comming from server with double slash so I used this:

var fixedText = yourString.replace(/(\r\n|\n|\r|\\n)/gm, '');

Try the following code. It works on all platforms.

var break_for_winDOS = 'test\r\nwith\r\nline\r\nbreaks';
var break_for_linux = 'test\nwith\nline\nbreaks';
var break_for_older_mac = 'test\rwith\rline\rbreaks';

break_for_winDOS.replace(/(\r?\n|\r)/gm, ' ');
'test with line breaks'

break_for_linux.replace(/(\r?\n|\r)/gm, ' ');
'test with line breaks'

break_for_older_mac.replace(/(\r?\n|\r)/gm, ' ');
// Output
'test with line breaks'


The easiest approach is using regular expressions to detect and replace newlines in the string. In this case, we use replace function along with string to replace with, which in our case is an empty string.

function remove_linebreaks( var message ) {
    return message.replace( /[\r\n]+/gm, "" );

In the above expression, g and m are for global and multiline flags

On mac, just use \n in regexp to match linebreaks. So the code will be string.replace(/\n/g, ''), ps: the g followed means match all instead of just the first.

On windows, it will be \r\n.