正则表达式删除块注释也删除 * 选择器 [英] Regex to remove block comments also removing * selector

问题描述

我正在尝试使用 bash 从 .css 文件中删除所有块注释.我有以下 sed 命令的正则表达式:

I am attempting to remove all block comments from .css files using bash.I have the following regex for a sed command:

sed -r '/\/(\*)*(\s)?(\w)*|(\*)(\s)?(\w)*/d'

这可以很好地去除块注释，例如:

This works fine stripping out block comments such as:

/**
* This is a comment
*/

/* this is another comment */

但它也在删除 * Selector 的实例

but it is also removing instances of the * Selector

* { 
     font-size: 10px;
     ...etc
  }

我怎样才能改变当前的正则表达式来考虑这种特殊情况?还是我需要重写我的正则表达式?

How could I alter the current regex to take this special case into account? Or will I need to rewrite my regex?

澄清一下，我将遇到的评论类型:

To clarify, the kinds of comments that I will encounter:

/**
* This is a comment
*/

/* This is another comment */

/* This is also 
valid
*/

/** "*/" as is { this } */

推荐答案

以下 sed 命令应该适合您:

The following sed command should work for you:

sed '/\/\*\**/{:a;/\*\//d;N;ba}' file.css

命令搜索模式 /** 或 /*，如果找到，则执行大括号 {cmd1;cmd2;...}.在该块中，我首先定义了一个标签 :a.在下一个命令中，我检查模式缓冲区是否包含结束的 */.如果找到，我将删除模式缓冲区并开始下一个循环.否则，我通过 N 将下一行输入附加到模式缓冲区，并通过 ba 转到标签 a.

The command searches for the pattern /** or /* and if it was found, it executes the block of commands between the curly braces {cmd1;cmd2;...}. Inside that block I first define a label :a. In the next command I check if the pattern buffer contains the closing */. If it was found I delete the pattern buffer and start the next cycle. Otherwise I append the next line of input to the patttern buffer through N and goto the label a through ba.

file.css:

/**
* This is a comment
*/
a {
    font-size: 10px;
}

/* This is another comment */
li {
    font-size: 12px;
}

/* This is also 
valid
*/
* {
    color:white;
}

/** "*/" as is { this } */
table {
    color:blue;
}

输出:

a {
    font-size: 10px;
}

li {
    font-size: 12px;
}

* {
    color:white;
}

table {
    color:blue;
}

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