JavaScript：如何使用正则表达式从字符串中删除空行？ [英] JavaScript: how to use a regular expression to remove blank lines from a string?

问题描述

我需要使用JavaScript删除HTML文本框中的空行。空行可以位于 textarea 元素中的任何位置。空行可以只是一个返回或空格加返回。

I need to use JavaScript to remove blank lines in a HTML text box. The blank lines can be at anywhere in the textarea element. A blank line can be just a return or white spaces plus return.

我期待一个正则表达式解决方案。以下是我试过的一些，但它们无法正常工作，无法找出原因：

I am expecting a regular expression solution to this. Here are some I tried, but they are not working and cannot figure out why:

/^\s*\r?\n/g   

/^\s*\r?\n$/g

编辑1

似乎aaronman和m建议的解决方案（我稍微修改了一下） .buettner的工作原理：

It appears that the solution (I modified it a little) suggested by aaronman and m.buettner works:

string.replace(/^\s*\n/gm, "") 

有人能说出为什么我的第一个正则表达式不起作用吗？

Can someone tell why my first regular expression is not working?

编辑2

阅读完所有有用的答案后，我想出了这个：

After reading all useful answers, I came up with this:

/^[\s\t]*(\r\n|\n|\r)/gm

这是否适用于所有情况？

Is this going to be one that cover all situations?

编辑3

这是最简洁的一个，涵盖所有空格（空格，标签）和平台（Linux，Windows，Mac）。

This is the most concise one covering all spaces (white spaces, tabs) and platforms (Linux, Windows, Mac).

/^\s*[\r\n]/gm

非常感谢m.buettner！

Many thanks to m.buettner!

推荐答案

您的模式似乎没问题，您只需要包含多行修饰符 m ，所以 ^ 和 $ 匹配行的开头和结尾：

Your pattern seems alright, you just need to include the multiline modifier m, so that ^ and $ match line beginnings and endings as well:

/^\s*\n/gm

如果没有 m ，锚点只匹配字符串开头和结尾。

Without the m, the anchors only match string-beginnings and endings.

请注意，你错过了在Mac行结尾（仅 \ r ）。在这种情况下，这将有所帮助：

Note that you miss out on Mac line endings (only \r). This would help in that case:

/^\s*[\r\n]/gm

另请注意（在这两种情况下）您不需要匹配可选的 \\ \\ r \\ n 明确地在 \ n 前面，因为这是由 \ * * 。

Also note that (in both cases) you don't need to match the optional \r in front of the \n explicitly, because that is taken care of by \s*.

Dex 指出注释，如果它只包含空格（并且后面没有换行符），则无法清除最后一行。解决这个问题的方法是使实际的换行符可选，但在它之前包含一个行尾锚点。在这种情况下，你做必须匹配正确结束的行，但是：

As Dex pointed out in a comment, this will fail to clear the last line if it consists only of spaces (and there is no newline after it). A way to fix that would be to make the actual newline optional but include an end-of-line anchor before it. In this case you do have to match the line ending properly though:

/^\s*$(?:\r\n?|\n)/gm

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