How do I remove trailing whitespace using a regular expression?

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.

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UPDATE (2016):

Nowadays I automate such code cleaning by using Sublime's [TrailingSpaces][1] package, with custom/user setting:

"trailing_spaces_trim_on_save": true

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

[1]: https://github.com/SublimeText/TrailingSpaces "TrailingSpaces"

Try just removing trailing spaces and tabs:

[ \t]+$

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

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

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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);

Regex to find trailing and leading whitespaces:

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

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

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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.

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

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

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.

To remove any blank trailing spaces use this:

\n|^\s+\n

I tested in the Atom and Xcode editors.

In Java:

 

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

You can simply use it like this:

var regex = /( )/g;

Sample: click here