How do I remove trailing whitespace using a regular expression?

Question

I want to remove trailing white spaces and tabs from my code without removing empty lines.

I tried:

\s+$

and:

([^\n]*)\s+\r\n

But they all removed empty lines too. I guess \s matches end-of-line characters too.

UPDATE (2016):

Nowadays I automate such code cleaning by using Sublime's TrailingSpaces package, with custom/user setting:

"trailing_spaces_trim_on_save": true

It highlights trailing white spaces and automatically trims them on save.

Your guess is correct. `\s` matches all kinds of whitespace characters. — Tim Pietzcker, Mar 02 '12 at 11:20
The corresponding *Vi and Vim* question is *[What's the simplest way to strip trailing whitespace from all lines in a file?](https://vi.stackexchange.com/questions/454)* — Peter Mortensen, Jul 05 '17 at 20:59
Here is a [doc entry](http://vim.wikia.com/wiki/Remove_unwanted_spaces) on topic. — , Aug 06 '18 at 08:40

score 171 · Accepted Answer · answered Mar 02 '12 at 11:15

171

Try just removing trailing spaces and tabs:

[ \t]+$

answered Mar 02 '12 at 11:15

qbert220

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How would I exclude lines which contain only whitespaces, tabs or a mixture of them? – Daniel F Jun 05 '13 at 10:14
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@DanielF. `([^ \t])[ \t]+$`, but you'd have to replace with `\1` instead of an empty string. – KOVIKO Jan 04 '14 at 13:49
@Koviko Thanks, but what I meant was that lines which contain only whitespaces or tabs, like indented functions in Python, where indentation is important, should be left as they are. I don't see the suggested regex doing that. – Daniel F Jan 05 '14 at 23:28
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@DanielF. The regex I posted looks for non-space characters preceding the space characters, and preserves the non-space characters with the `\1` replacement. Lines that consist of only space characters will not match this regex and will be skipped. – KOVIKO Jan 07 '14 at 04:06
@Koviko This should showcase the issue: http://jsfiddle.net/s7vW5/2/ .The whitespace-only line between foo and bar should contain two spaces. Python's flavor of regex behaves the same. – Daniel F Jan 07 '14 at 12:15
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@DanielF. In my case, I was using a text editor (Notepad++) that counts each line as a string, not the entire document as a string. As such, newline characters are ignored. To handle the case where newline characters are not ignored, simply add them to the regex: `([^ \t\r\n])[ \t]+$` – KOVIKO Jan 08 '14 at 15:30
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@Koviko: Since your comment is the correct solution to the problem and the answer you've commented on is not, I suggest you post that regex as a separate answer. Your comment is valuable, but easy to miss. – Slipp D. Thompson Jan 28 '14 at 00:50
For Atom, it's `[ \t]+$` – Zhanwen Chen Jul 06 '21 at 18:25

KOVIKO · Answer 2 · 2018-05-29T01:34:12.483

38

To remove trailing whitespace while also preserving whitespace-only lines, you want the regex to only remove trailing whitespace after non-whitespace characters. So you need to first check for a non-whitespace character. This means that the non-whitespace character will be included in the match, so you need to include it in the replacement.

Regex: ([^ \t\r\n])[ \t]+$

Replacement: \1 or $1, depending on the IDE

edited May 29 '18 at 01:34

answered Jan 29 '14 at 04:13

KOVIKO

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Awesome. Thank you. Also, the same regex technique modified to use a lookbehind assertion: `(?<![ \t\r\n])[ \t]+$` (replacement pattern should just be blank). – Slipp D. Thompson Mar 03 '15 at 20:21
What platform uses "\1" as replacement? – Peter Mortensen May 31 '15 at 15:13
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@PeterMortensen Notepad++, for one. – KOVIKO May 31 '15 at 15:14
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@PeterMortensen `sed` too. – ocodo May 27 '18 at 13:11
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This is a built-in function in Notepad++: Menu *Edit* → *Blank Operations* → *Trim Trailing Space* – Peter Mortensen Jul 03 '19 at 23:26
@PeterMortensen That will not preserve whitespace-only lines, as in the first line of my answer. – KOVIKO Jul 05 '19 at 14:45
None of the answers here appear to handle when the text before the trailing spaces contains spaces itself, e.g. "E:\Data\Shared\SCANNED DOCUMENTS " – Rob Nicholson Dec 24 '22 at 10:53

score 19 · Answer 3 · edited Jun 20 '20 at 09:12

19

The platform is not specified, but in C# (.NET) it would be:

Regular expression (presumes the multiline option - the example below uses it):

    [ \t]+(\r?$)

Replacement:

$1

For an explanation of "\r?$", see Regular Expression Options, Multiline Mode (MSDN).

Code example

This will remove all trailing spaces and all trailing TABs in all lines:

string inputText = "     Hello, World!  \r\n" +
                   "  Some other line\r\n" +
                   "     The last line  ";
string cleanedUpText = Regex.Replace(inputText,
                                     @"[ \t]+(\r?$)", @"$1",
                                     RegexOptions.Multiline);

edited Jun 20 '20 at 09:12

Community

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answered May 31 '15 at 15:22

Peter Mortensen

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The regular expressions in the other two answers do not work in .NET (they don't result in any replacement - the text is left unchanged). – Peter Mortensen Jun 01 '15 at 17:11
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This also works in Xcode and Swift. Thumbs way up. Thank you. – ScottyBlades Mar 07 '21 at 08:18

score 4 · Answer 4 · edited Mar 18 '19 at 13:58

4

Regex to find trailing and leading whitespaces:

^[ \t]+|[ \t]+$

edited Mar 18 '19 at 13:58

Peter Mortensen

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answered Jan 18 '18 at 06:06

mohanish

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Can you provide some context? Where and how did you test it? – Peter Mortensen Mar 04 '20 at 19:43

score 3 · Answer 5 · edited Mar 18 '19 at 13:59

If using Visual Studio 2012 and later (which uses .NET regular expressions), you can remove trailing whitespace without removing blank lines by using the following regex

Replace (?([^\r\n])\s)+(\r?\n)

With $1

Some explanation

The reason you need the rather complicated expression is that the character class \s matches spaces, tabs and newline characters, so \s+ will match a group of lines containing only whitespace. It doesn't help adding a $ termination to this regex, because this will still match a group of lines containing only whitespace and newline characters.

You may also want to know (as I did) exactly what the (?([^\r\n])\s) expression means. This is an Alternation Construct, which effectively means match to the whitespace character class if it is not a carriage return or linefeed.

Alternation constructs normally have a true and false part,

(?( expression ) yes | no )

but in this case the false part is not specified.

score 3 · Answer 6 · edited Nov 17 '21 at 22:46

3

[ |\t]+$ with an empty replace works.

\s+($) with a $1 replace also works, at least in Visual Studio Code...

edited Nov 17 '21 at 22:46

Mark

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answered Dec 10 '18 at 21:36

Charles Goodwin

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score 0 · Answer 7 · edited Jun 09 '16 at 08:33

0

To remove trailing white space while ignoring empty lines I use positive look-behind:

(?<=\S)\s+$

The look-behind is the way go to exclude the non-whitespace (\S) from the match.

edited Jun 09 '16 at 08:33

Andreas

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answered Apr 16 '16 at 23:46

Markus Strauss

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This matches empty lines. The match will *start* at the end of a non-empty line, but it will include any empty lines that follow. – Alan Moore Jun 09 '16 at 09:31
What platform did you use? Perl? – Peter Mortensen Jun 09 '16 at 11:39

score 0 · Answer 8 · edited Jun 26 '19 at 23:24

0

To remove any blank trailing spaces use this:

\n|^\s+\n

I tested in the Atom and Xcode editors.

edited Jun 26 '19 at 23:24

Peter Mortensen

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answered Apr 09 '19 at 10:46

Oskar

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What replacement character did you use? `$1`? Or something else? – Peter Mortensen Mar 07 '21 at 21:10
An empty string as the replacement? – Peter Mortensen Apr 28 '21 at 21:52

score 0 · Answer 9 · answered Mar 28 '23 at 14:00

0

Strangely, I could not get the /[\t ]+$/m to work in PHP. I tried variants of the other answers here, but still failed.

So, I went with this replacement instead:

<?php
$contents = preg_replace('/[\t ]+(\v)/', '$1', $contents);

answered Mar 28 '23 at 14:00

Ryan Briscall

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score -1 · Answer 10 · answered Jan 29 '19 at 06:33

-1

In Java:



String str = "    hello world  ";

// prints "hello world" 
System.out.println(str.replaceAll("^(\\s+)|(\\s+)$", ""));

answered Jan 29 '19 at 06:33

john smith

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The OP didn't want to remove leading spaces. – Armali Jan 29 '19 at 06:55

score -4 · Answer 11 · edited Mar 18 '19 at 14:04

-4

You can simply use it like this:

var regex = /( )/g;

Sample: click here

edited Mar 18 '19 at 14:04

Peter Mortensen

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answered Feb 12 '19 at 05:17

Kumaresan Sd

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Hi @PeterMortensen, it will remove all white spaces – Kumaresan Sd Mar 19 '19 at 05:22
And what was the question? – Peter Mortensen Mar 07 '21 at 21:09

How do I remove trailing whitespace using a regular expression?

11 Answers11

Code example

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