撼动正则表达式天地?!

[oicqkill]

谁能给出下面正确的解呢
如下的代码，如何用正则表达式正确的去除注释:
var kk = "// sljdfljsdf"; // *skldflsfllk"sdlf//sdfdsf"
var kk = "// sljdfljsdf"; // *skldflsfllk"sdlf//sdfdsf"
var kk = "// sljdfljsdf"; // *skldflsfllk"sdlf//sdfdsf"
var kk = "// sljdf/*sdf*/ljsdf'; // *sk/*fdsf*/ld""flsfllk"sdlf//sdfdsf"
var kk = '// sljdf/*sdf*/ljsdf'; // *sk/*fdsf*/ld""flsfllk"sdlf//sdfdsf"
var kk = '// sljdf/*sdf*/ljsdf'; // *sk/*fdsf*/ld""flsfllk"sdlf//sdfdsf"

[火龙果被占用了]

不明白！// 后面的不都是注释吗？

[oicqkill]

就是把这段代码做一个串来处理，用正则表达式正确去除注释
例如:
var kk = "// sljdfljsdf"; // *skldflsfllk"sdlf//sdfdsf"
var kk = "// sljdfljsdf"; // *skldflsfllk"sdlf//sdfdsf"
var kk = "// sljdfljsdf"; // *skldflsfllk"sdlf//sdfdsf"
var kk = "// sljdf/*sdf*/ljsdf' // *sk/*fdsf*/ld""flsfllk"sdlf//sdfdsf"
var kk = '// sljdf/*sdf*/ljsdf'; // *sk/*fdsf*/ld""flsfllk"sdlf//sdfdsf"
var kk = '// sljdf/*sdf*/ljsdf' // *sk/*fdsf*/ld""flsfllk"sdlf//sdfdsf"
处理后为:
var kk = "// sljdfljsdf";
var kk = "// sljdfljsdf";
var kk = "// sljdfljsdf";
var kk = "// sljdf/*sdf*/ljsdf'
var kk = '// sljdf/*sdf*/ljsdf';
var kk = '// sljdf/*sdf*/ljsdf'

[LexChen]

关键是"的处理，呵呵

[swvincent]

kk.replace(/^\/\/\\*w+$/,'');
不知道行不！！！

[windstore]

正则表达式如下：
//(?<=;\s*//)[^\n]+

[windstore]

kk.replace(/\/\/(?<=;\s*//)[^\n]+/,'');

[碧海情天-赵亮]

感觉是很闲的问题

[○福泽天下○]

不会

[qiuruiok]

呃....我什么都不懂...

[shan1119]

楼上的就可以,以分号为分界点.

[believefym]

js不支持反向预搜索

[xinxin821103]

不懂,顶一下.

[xiaoniaoxiangfei]

我想问一下，弄这个有什么用吗？

[LexChen]

净化代码

[yingzi107]

程序员群 41801837

[CrazyGou]

<span id="text" style="display:none">
<pre>
var kk = "// sljdfljsdf"; // *skldflsfllk"sdlf//sdfdsf"
var kk = "// sljdfljsdf"; // *skldflsfllk"sdlf//sdfdsf"
var kk = "// sljdfljsdf"; // *skldflsfllk"sdlf//sdfdsf"
var kk = "// sljdf/*sdf*/ljsdf' // *sk/*fdsf*/ld""flsfllk"sdlf//sdfdsf"
var kk = '// sljdf/*sdf*/ljsdf'; // *sk/*fdsf*/ld""flsfllk"sdlf//sdfdsf"
var kk = '// sljdf/*sdf*/ljsdf' // *sk/*fdsf*/ld""flsfllk"sdlf//sdfdsf"
var kk = new Object(); //test
</pre>
</span>

<script>
var reg = /([^'"\/\n]*(['"]?).*?[^\\]\2(?:\s*;\s*|\s+))(?:\/\/.*|\/\*.*\*\/\s*)(\n|$)/g;
var kk = document.getElementById("text").innerText;
alert("处理前：\n" + kk);
alert("处理后：\n" + kk.replace(reg, "$1$3"));
</script>

[嗷嗷叫的老马]

.........................................

[javajapanengineer]

http://community.csdn.net/Expert/topic/5541/5541735.xml?temp=.7890283
参考一下~~~~~~~~

[oicqkill]

非常抱歉，以上回答都还没有满足需求。

[a_nuo]

用java实现可以用正则表达式\s+\/\/\s+\*.+\/\/\w+\"

import java.util.regex.Pattern;
import java.util.regex.Matcher;
import java.util.Arrays;

public class TestRegex {
        
    /**
     * Creates a new instance of <code>TestRegex</code>.
     */
    public TestRegex() {
    }
    
    /**
     * @param args the command line arguments
     */
    public static void main(String[] args) {
    	String regEx="\\s+\\/\\/\\s+\\*.+\\/\\/\\w+\"";
    	String oldCode="var   kk   =   \"//   sljdfljsdf\";   //   *skldflsfllk\"sdlf//sdfdsf\"\r\nvar   kk   =   \"//   sljdfljsdf\";   //   *skldflsfllk\"sdlf//sdfdsf\"\r\nvar   kk   =   \"//   sljdfljsdf\";   //   *skldflsfllk\"sdlf//sdfdsf\"\r\nvar   kk   =   \"//   sljdf/*sdf*/ljsdf \";   //   *sk/*fdsf*/ld\"\"flsfllk\"sdlf//sdfdsf\"\r\nvar   kk   =   \"//   sljdf/*sdf*/ljsdf \";   //   *sk/*fdsf*/ld\"\"flsfllk\"sdlf//sdfdsf\"\r\nvar   kk   =   \"//   sljdf/*sdf*/ljsdf \";   //   *sk/*fdsf*/ld\"\"flsfllk\"sdlf//sdfdsf\"";
        Pattern patter=Pattern.compile(regEx);
        Matcher m=patter.matcher(oldCode);
        StringBuffer newCode=new StringBuffer();
        while (m.find()) {
        	System.out.println(m.group());
        	m.appendReplacement(newCode,"");
        }
        m.appendTail(newCode);
        System.out.println("oldCode:"+oldCode+"\nNewCode:");
        System.out.println(newCode);
    	
        // TODO code application logic here
    }
}

试过了结果真确