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4.极验滑块 跟W值3
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@ -219,4 +219,81 @@
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r[$_BEGJJ(444)]($_BEGIH(33)) + $_BEGIH(407) + i[$_BEGIH(444)]($_BEGJJ(33)) + $_BEGIH(407) + o[$_BEGIH(444)]($_BEGIH(33));
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r[$_BEGJJ(444)]($_BEGIH(33)) + $_BEGIH(407) + i[$_BEGIH(444)]($_BEGJJ(33)) + $_BEGIH(407) + o[$_BEGIH(444)]($_BEGIH(33));
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}
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}
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发现是平坦流,线开头结尾断点
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最轨迹值的理解
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发现是平坦流,线开头结尾断点,发现这里面有一个t值,非常像轨迹值
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打断点到最后一行,发现拼接好的轨迹
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r[$_BEGJJ(444)]($_BEGIH(33)) + $_BEGIH(407) + i[$_BEGIH(444)]($_BEGJJ(33)) + $_BEGIH(407) + o[$_BEGIH(444)]($_BEGIH(33));
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进行字符串还原
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r["join"]("") + "!!" + i["join"]("") + "!!" + o["join"]("");
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这里就涉及到r, i, o,这里通过全局函数的方式调用js代码,通过传入轨迹值生成加密数据
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var t = function(t) {
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var $_BEHDi = lTloj.$_CX
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, $_BEHCK = ['$_BEHGM'].concat($_BEHDi)
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, $_BEHEF = $_BEHCK[1];
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$_BEHCK.shift();
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var $_BEHFx = $_BEHCK[0];
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for (var e, n, r, i = [], o = 0, s = 0, a = t[$_BEHEF(182)] - 1; s < a; s++)
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e = Math[$_BEHEF(156)](t[s + 1][0] - t[s][0]),
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n = Math[$_BEHDi(156)](t[s + 1][1] - t[s][1]),
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r = Math[$_BEHDi(156)](t[s + 1][2] - t[s][2]),
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0 == e && 0 == n && 0 == r || (0 == e && 0 == n ? o += r : (i[$_BEHEF(140)]([e, n, r + o]),
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o = 0));
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return 0 !== o && i[$_BEHDi(140)]([e, n, o]),
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i;
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}(this[$_BEGJJ(361)])
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这个的 this[$_BEGJJ(361)] 就是轨迹数据,我通过传值的方式改造成
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js末尾进行全局函数引出
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window.guiji = W[$_CJET(261)]["\u0024\u005f\u0047\u0045\u0079"];
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代码调试
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同样轨迹代码也是一样炮制
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l = n["$_CIBw"]["$_BBCA"](n["$_CIBw"]["$_GEy"](), n["$_CIY"]["c"], n["$_CIY"]["s"]);
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window.guiji_2 = W[$_CJET(261)]["\u0024\u005f\u0042\u0042\u0043\u0041"];
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走到这里 aa 的值就得到了,接下来拿到 l 值
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l = V[$_CAGEe(342)](gt[$_CAGEe(209)](o), r[$_CAGEe(742)]())
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window.get_v = V["encrypt"];
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最后一葫芦滑瓢,w值得到
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var u = r[$_CAGEe(750)]()
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, l = V[$_CAGEe(342)](gt[$_CAGEe(209)](o), r[$_CAGEe(742)]())
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, h = m[$_CAGEe(733)](l)
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, f = {
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"\u0067\u0074": i[$_CAGEe(147)],
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"\u0063\u0068\u0061\u006c\u006c\u0065\u006e\u0067\u0065": i[$_CAGDp(154)],
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"\u006c\u0061\u006e\u0067": o[$_CAGDp(119)],
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"\u0024\u005f\u0042\u0042\u0046": r[$_CAGEe(623)],
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"\u0063\u006c\u0069\u0065\u006e\u0074\u005f\u0074\u0079\u0070\u0065": r[$_CAGEe(648)],
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"\u0077": h + u
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};
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注意点:如果函数出现this,那就要导出对象,而不是导出方法。
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