暴力破解带有密码的压缩文件

全栈程序员-用户IM • 2022年5月7日下午6:00 • 未分类

大家好，又见面了，我是你们的朋友全栈君。

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。
暴力破解带有密码的压缩文件
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。

发布者：全栈程序员-用户IM，转载请注明出处：https://javaforall.cn/139685.html原文链接：https://javaforall.cn

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