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