# 逆向百度翻译的一次小尝试

## 逆向步骤

3. 算出sign,根据查询入参，js逆向算出sign

## 具体步骤

3.在控制台中搜索 sign: 找到查询的js, 抠出I(e)函数生成加密的sign方法

js sign代码

function n(r, o) {
for (var t = 0; t < o.length - 2; t += 3) {
var a = o.charAt(t + 2);
a = a >= "a" ? a.charCodeAt(0) - 87 : Number(a),
a = "+" === o.charAt(t + 1) ? r >>> a : r << a,
r = "+" === o.charAt(t) ? r + a & 4294967295 : r ^ a
}
return r
};

function e(r) {
var o = r.match(/[\uD800-\uDBFF][\uDC00-\uDFFF]/g);

var i = null;

if (null === o) {
var t = r.length;
t > 30 && (r = "" + r.substr(0, 10) + r.substr(Math.floor(t / 2) - 5, 10) + r.substr(-10, 10))
} else {
for (var e = r.split(/[\uD800-\uDBFF][\uDC00-\uDFFF]/), C = 0, h = e.length, f = []; h > C; C++)
"" !== e[C] && f.push.apply(f, a(e[C].split(""))),
C !== h - 1 && f.push(o[C]);
var g = f.length;
g > 30 && (r = f.slice(0, 10).join("") + f.slice(Math.floor(g / 2) - 5, Math.floor(g / 2) + 5).join("") + f.slice(-10).join(""))
}
var u = void 0
, l = "" + String.fromCharCode(103) + String.fromCharCode(116) + String.fromCharCode(107);
u = null !== i ? i : (i = "320305.131321201" || "") || "";
for (var d = u.split("."), m = Number(d[0]) || 0, s = Number(d[1]) || 0, S = [], c = 0, v = 0; v < r.length; v++) {
var A = r.charCodeAt(v);
128 > A ? S[c++] = A : (2048 > A ? S[c++] = A >> 6 | 192 : (55296 === (64512 & A) && v + 1 < r.length && 56320 === (64512 & r.charCodeAt(v + 1)) ? (A = 65536 + ((1023 & A) << 10) + (1023 & r.charCodeAt(++v)),
S[c++] = A >> 18 | 240,
S[c++] = A >> 12 & 63 | 128) : S[c++] = A >> 12 | 224,
S[c++] = A >> 6 & 63 | 128),
S[c++] = 63 & A | 128)
}
for (var p = m, F = "" + String.fromCharCode(43) + String.fromCharCode(45) + String.fromCharCode(97) + ("" + String.fromCharCode(94) + String.fromCharCode(43) + String.fromCharCode(54)), D = "" + String.fromCharCode(43) + String.fromCharCode(45) + String.fromCharCode(51) + ("" + String.fromCharCode(94) + String.fromCharCode(43) + String.fromCharCode(98)) + ("" + String.fromCharCode(43) + String.fromCharCode(45) + String.fromCharCode(102)), b = 0; b < S.length; b++)
p += S[b],
p = n(p, F);
return p = n(p, D),
p ^= s,
0 > p && (p = (2147483647 & p) + 2147483648),
p %= 1e6,
p.toString() + "." + (p ^ m)
};

e('工作');

github.com/Xuzan9396/my_baidufanyi

1周前 评论

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