在bash中删除多行降价评论

Question

如何在bash中删除多行降价评论，例如下面的评论？

some text

<!-- QUESTION:
How do I remove everything
in-between these tags? 
 -->

some<!-- Including embedded single-line comments such as this --> text

我尝试了sed -e 's/<!--((.*?)\n?)+-->//g' $1，它仅适用于单行，而cat $1 | tr '\n' '\r' | sed -e 's/<!--.*-->//g' | tr '\r' '\n'则删除了第一个多行注释后的所有内容。

<!--((.*?)\n?)+-->在我的文本编辑器中捕获所需的区域，但是 sed -e 's/<!--((.*?)\n?)+-->//g' $1无效。

我可以发现使用C ++注释的其他示例太复杂，无法解码。

Answer 1

您可以使用perl one-liner完成此操作。

Perl切换：

-0 sets the input record separator to the null character \0
-p prints the result of perl code
-e executes the following code

正则表达式：

g flag means global (perform the replacement as many times as possible)
s flag means treat the input as a multi-line string

Match the characters `<!--` followed by anything up to the characters `-->`
including anything after that till the newline.  Replace that with nothing. 

在行动：

perl -0pe 's|<!--.+?-->.*?\n||gs;' input

<强>输出：

some text


some text