暴力破解带有密码的压缩文件

墨蓝 2023-07-10 08:25 28阅读 0赞

一 **暴力破解**，就是不断生成新的密码去尝试能否破解成功。假如我们的密码位数为3位，其里面的值有可能是0或者1，其产生的密码集的数量为2的3次幂，也就是八个，然后一个一个去破解。而这生成这8个密码的过程就有一个递归规则存在。  
![watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzMwNTUzNzcz_size_16_color_FFFFFF_t_70][]  
我们传入深度deep和对应字符串passwd两个参数，如果深度为1，直接输出字符串passwd，如果深度不为1，把深度deep减一，并把可能存在的字符串集合\[0,1\]遍历,添加到passwd的后面，重新回到这个递归函数，直到深度deep为1输出。

二 解压压缩文件  
网上有对应的jar包去解压压缩文件，但是我这里使用一种比较简单的方法去实现，本地电脑window上装有winrar，java调用本地winrar.exe去解压压缩文件。  
其核心代码为：

C:\Program Files\WinRAR\WinRAR.exe  x -ibck -hp123 -y F:\\BaiduNetdiskDownload\\yewen4\\a.rar D:\\test\\unrar\\

第一个路径是winrar.exe的路径，第二个路劲是需要解压缩文件的路径地址，第三个是存放的路径。-hp后面的123表示密码。

下面是代码的实现

/**
         * 
         * @param filePath 解压文件路径
         * @param targetPath 解压之后存放的路径
         * @param passwd 密码
         * @return
         * @throws Exception
         */
    public static int unrar(String filePath,String targetPath,String passwd) throws Exception {
    		//winrar的执行路径
            String rarPath="C:\\Program Files\\WinRAR\\WinRAR.exe";
            StringBuilder sb = new StringBuilder();
            sb.append(rarPath);
            sb.append(" x -ibck -hp");
            sb.append(passwd).append(" -y ").append(filePath+" "+targetPath);
            Process process;
    
            process = Runtime.getRuntime().exec(sb.toString());
            if(process.waitFor() ==0 ){
                FileOutputStream fileOutputStream = new FileOutputStream(new File("D:\\test\\unrar\\getPassyewen4.txt"));
                String s = "解压后的密码："+passwd;
                fileOutputStream.write(s.getBytes());
                fileOutputStream.flush();
                fileOutputStream.close();
                return 1;
            }else{
                System.out.println(new Date()+"--失败："+passwd);
                return 0 ;
            }
        }
    
    	 /**
         * 递归获取密码过程
         * @param deep 深度
         * @param parent 密码
         * @return
         * @throws Exception
         */
        public static int findpwd(int deep,String parent) throws Exception {
            String[] dir ={"q","w","e","r","t","y","u","i","o","p","a","s","d","f","g","h","j","k","l",
                    "z","x","c","v","b","n","m","Q","W","E","R","T","Y","U","I","O","P","A","S","D","F","G","H","J","K","L","Z","X",
                    "C","V","B","N","M","1","2","3","4","5","6","7","8","9","0","`","!","@","#","$","%","&","*","(",")","-","_","+","=","[",
                    "]","{","}","|","\\","\"",":",";","\'","<",">",",",".","?","/"};
            if(deep == 1){
                if(unrar("","",parent) == 1){
                    return 1;
                }else{
                    return 0;
                }
            }else{
                for(int j =0;j<dir.length;j++){
                    if(findpwd(deep-1,parent+dir[j]) ==1 ){
                        return 1;
                    }
                }
            }
            return 0;
        }

文章最后的总结：  
破解的难度跟设置密码的长度有关，破解一位的密码只需要92的1次幂个密码，而破解六位的密码需要92的6次幂个密码，其难度可想而知。所以这里写的内容只供学习参考。

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