java离线日期授权解决方案 解决改系统日期 Java代码混淆 日期混淆

在java项目中,需要客户定制离线授权时,怎么保证软件时效性呢?  如果用日期授权比如一年,如果单纯的获取系统时间,在离线的情况下,如果客户修改系统时间,软件的授权无法控制. 导致经济利益的损失.

下面我介绍写我的方案: 

1.没有做授权方案的代码, 修改系统时间,反编译代码, 所有逻辑一览无余,无法保证内容不被破解,复制,修改

package com.haiteng.action.autoreport;
 
import java.text.DateFormat;
import java.text.SimpleDateFormat;
import java.util.Date;
/**
 * 
 * ClassName:TestStrings
 * Description:测试 字符串加密方案
 * author: Q/V:75238016
 * date:2021年11月15日 - 下午5:11:08
 */
public class TestStrings {
    
    //加密串
    public static String strpw= "888888888888";
    //授权日期
    public static String strDate= "2022-08-08 08:08:08";
    //测试
    public static void main(String[] args) throws Exception {
        DateFormat df = new SimpleDateFormat("yyyy-MM-dd hh:mm");
        //判断当前密码是否一致
        if("888888888888".equals(strpw)) {
            //执行相关操作
            System.out.println("密码验证成功...");
        }
        //获取当前时间
        Date date = new Date();
        //判断当前过期时间
        if(df.parse(strDate).getTime() > date.getTime()) {
            //执行相关操作
            System.out.println("时间验证通过...");
        }
    }
}

2.采用混淆方法,内置时钟方案, 不用获取系统时间, 无法通过修改系统时间延长使用时间, 也无法反编译代码获取代码授权信息

package com.haiteng.action.autoreport;
 
import java.io.PrintStream;
import java.nio.charset.StandardCharsets;
import java.security.MessageDigest;
import java.text.DateFormat;
import java.text.SimpleDateFormat;
import java.util.Arrays;
import java.util.Base64;
import java.util.Base64.Decoder;
import java.util.Date;
import javax.crypto.Cipher;
import javax.crypto.spec.SecretKeySpec;
 
public class TestStrings
{
  public static void main(String[] NLCC)
    throws Exception
  {
    String JLCC;
    boolean kLCC;
    DateFormat mLCC = new SimpleDateFormat(A[b[2]]);
    if (G(A[b[3]].equals(strpw)))
      System.out.println(A[b[4]]);
    Date LLCC = new Date();
    if (!(h(f(mLCC.parse(strDate).getTime(), LLCC.getTime()))))
      return;
    System.out.println(A[b[5]]);
  }
 
  private static boolean h(int ???)
  {
    byte AJCC;
    return (??? > 0);
  }
 
  private static String b(String EkCC, String dkCC)
  {
    double wJCC;
    String AkCC;
    Exception bkCC;
    char CkCC;
    char dkCC;
    char dkCC;
    EkCC = new String(Base64.getDecoder().decode(EkCC.getBytes(StandardCharsets.UTF_8)), StandardCharsets.UTF_8);
    StringBuilder hkCC = new StringBuilder();
    char[] GkCC = dkCC.toCharArray();
    int fkCC = b[0];
    short ZJCC = EkCC.toCharArray();
    boolean yJCC = ZJCC.length;
    Exception XJCC = b[0];
    while (J(XJCC, yJCC))
    {
      char LkCC = ZJCC[XJCC];
      "".length();
      ++fkCC;
      ++XJCC;
      "".length();
      if (null != null)
        return null;
    }
    return String.valueOf(hkCC);
  }
 
  static
  {
    E();
    d();
    strpw = A[b[0]];
    strDate = A[b[1]];
  }
 
  private static String C(String oJCC, String LJCC)
  {
    short JJCC;
    byte kJCC;
    boolean LJCC;
    boolean LJCC;
    try
    {
      SecretKeySpec RJCC = new SecretKeySpec(MessageDigest.getInstance("MD5").digest(LJCC.getBytes(StandardCharsets.UTF_8)), "Blowfish");
      Cipher qJCC = Cipher.getInstance("Blowfish");
      qJCC.init(b[2], RJCC);
      return new String(qJCC.doFinal(Base64.getDecoder().decode(oJCC.getBytes(StandardCharsets.UTF_8))), StandardCharsets.UTF_8);
    }
    catch (Exception PJCC)
    {
      PJCC.printStackTrace();
    }
    return null;
  }
 
  private static String A(String ALCC, String ZkCC)
  {
    char VkCC;
    Exception wkCC;
    String ZkCC;
    String ZkCC;
    try
    {
      SecretKeySpec dLCC = new SecretKeySpec(Arrays.copyOf(MessageDigest.getInstance("MD5").digest(ZkCC.getBytes(StandardCharsets.UTF_8)), b[7]), "DES");
      Cipher CLCC = Cipher.getInstance("DES");
      CLCC.init(b[2], dLCC);
      return new String(CLCC.doFinal(Base64.getDecoder().decode(ALCC.getBytes(StandardCharsets.UTF_8))), StandardCharsets.UTF_8);
    }
    catch (Exception bLCC)
    {
      bLCC.printStackTrace();
    }
    return null;
  }
 
  private static void d()
  {
    A = new String[b[6]];
    A[b[0]] = A("qHu8wLLDF3GtZHy0018HAw==", "tVOKu");
    A[b[1]] = b("f0F4QHh9SWdCbW1BckhldUt6Sg==", "MqJrU");
    A[b[2]] = C("uX8Ds2zyr8EqcR6UE9pyWlqyUMsAaDgO", "VGtVr");
    A[b[3]] = b("QlJbXE5CUltcTkJS", "zjcdv");
    A[b[4]] = A("JktDF9fl3OwDkhnixRjda/Z9EUCprgtJ", "SfSkl");
    A[b[5]] = A("JTPA6jQSVMwNKBts2Mfou3sEhcAMbgpN", "JMCOI");
  }
 
  private static boolean G(int ???)
  {
    float CJCC;
    return (??? != 0);
  }
 
  private static void E()
  {
    b = new int[8];
    b[0] = ((36 + 67 - -22 + 6 ^ 67 + 63 - -11 + 38) & (0xEC ^ 0xC3 ^ 0x8C ^ 0x93 ^ -" ".length()));
    b[1] = " ".length();
    b[2] = "  ".length();
    b[3] = "   ".length();
    b[4] = (0x1E ^ 0x1A);
    b[5] = (0x6 ^ 0x3);
    b[6] = (0xC3 ^ 0xC5);
    b[7] = (47 + 112 - 69 + 76 ^ 164 + 167 - 330 + 173);
  }
 
  private static int f(long paramLong1, long paramLong2)
  {
    return (paramLong1 < paramLong2);
  }
 
  private static boolean J(int ???, int arg1)
  {
    int i;
    byte EJCC;
    return (??? < i);
  }
}

需要方案,制定化的,欢迎大家交流学习！

qq/wx  : 75238016