滑快破解2

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luzhisheng 2022-01-26 17:54:29 +08:00
parent 6631ef5797
commit d2dd9bc109
3 changed files with 8449 additions and 1 deletions

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@ -160,6 +160,10 @@
![debugger](../img/20.png)
最后的条件运算代码
return 0 == (1 & r[$_JJJB(182)]) ? r : $_JJJB(44) + r;
条件运算符是唯一的三元运算符,其语法格式如下:
b ? x : y
@ -215,4 +219,61 @@
return 0 == (1 & r["length"]) ? r : "0" + r;
}
发现这是一断算法代码,接下来的思路
1.扣出逆向代码,
2.既然已经知道rsa加密那么找到公钥
// rsa 大概流程
// 公钥
var publicKey = "MIGfMA***************AB";
// 创建一个对象
var encrypt = new JSEncrypt();
// 设置公钥
encrypt.setPublicKey(publicKey);
// 加密
var v_str = encrypt.encrypt(v_str); //需要加密的内容
console.log(v_str);
console.log(stringToHex(v_str));
alert(v_str);
// 私钥解密数据
var privateKey="MIIC*****qLNnis=";
var decrypt = new JSEncrypt();
decrypt.setPrivateKey(privateKey);
//解密数据
var uncrypted = decrypt.decrypt(v_str);//需要解密的内容
console.log(uncrypted);
## 将内部函数改变成全局变量
./zy-极验滑快/破解第一个参数u.js
调用
![debugger](../img/22.png)
window.ayf
ƒ E() {
var $_DBDBL = lTloj.$_DP()[2][4];
for (; $_DBDBL !== lTloj.$_DP()[2][3];) {
switch ($_DBDBL) {
case l…
var e = new window.ayf()["encrypt"]("c4d8355ea7a04436");
undefined
e
"8a140d60d5d3b94672c97a0150a23538e8fa1fd37370f89ce399ce96f756a6a3b61766f9bac4abcaa7aad132a6407d104a9de4c91a6a25a5ecd5d9a95da362c0c9dfb8351b40392e250f0c58a98363d4d622e38cdbd6f2e6ff8dfd9de94a2a372e4e957d5bae446f1226279dac6ed672fafd1578d462fb23fbee55653acfac4e"
最终调用方案
function get_id(){
function random_(){
var data= "";
for(var i=0;i<4;i++){
data += (65536 * (1 + Math["random"]()) | 0)["toString"](16)["substring"](1)
}
return data
}
return new window.ayf()["encrypt"](random_());
}

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