如何使用正则表达式删除尾随空格？

我想从我的代码中移除尾随的空格和制表符 删除空行。

我试过:

\s+$

以及:

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

但是他们都删除了空行。我猜 \s也匹配行尾字符。

更新(2016) :

现在，我通过使用 Sublime 的 尾随空间软件包自动清除这些代码，具有自定义/用户设置:

"trailing_spaces_trim_on_save": true

它突出显示尾部的空白，并在保存时自动修剪它们。

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最佳答案

Try just removing trailing spaces and tabs:

[ \t]+$
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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

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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);
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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.

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Regex to find trailing and leading whitespaces:

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

Enter image description here

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.

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

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

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In Java:




String str = "    hello world  ";


// prints "hello world"
System.out.println(str.replaceAll("^(\\s+)|(\\s+)$", ""));
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You can simply use it like this:

var regex = /( )/g;

Sample: click here

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To remove any blank trailing spaces use this:

\n|^\s+\n

I tested in the Atom and Xcode editors.


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