js逆向解析之百度翻译案例

百度翻译使用了js函数对sign值进行加密，所以需要用到js逆向解析

js逆向解析即通过调用对方的js加密函数来获取正确的加密数据

js逆向解析需要安装js开发环境node.js和execjs模块

有两个文件，一个py文件存放主程序，和一个js文件存放从浏览器中找到的生成js加密数据的js代码
两个文件需在同一个文件夹中。

*.py py文件

import requests    # 需安装
import execjs      # pip install pyexecjs 用于运行js代码
import jsonpath    # 需安装
import time

class BaiduFanyi:
    def __init__(self):
        self.url = 'https://fanyi.baidu.com/v2transapi?from=zh&to=en'
        self.head = {
            "user-agent": "",       # 请到浏览器开发者工具中复制
            "referer": "https: // fanyi.baidu.com /",
            "cookie": "",       # 请到浏览器开发者工具中复制
        }
        self.zh = None
        self.sign = None
        self.formdata = {"from": "zh",
                         "to": "en",
                         "query": self.zh,
                         "transtype": "enter",
                         "simple_means_flag": "3",
                         "sign": self.sign,
                         "token": "",      # 请到浏览器开发者工具中复制
                         "domain": "common"}

    def get_sign(self):
        # 1、打开js文件，使用js代码生成js加密数据
        with open('sign.js', 'r') as f:
            js_sign = f.read()
        # 2、编译
        exe_sign = execjs.compile(js_sign) 
        # 3、调用js函数生成加密数据
        self.sign = exe_sign.call('e', self.zh)

    def fanyi(self, zh):
        # 翻译
        self.zh = zh
        self.get_sign()
        self.formdata['query'] = self.zh
        self.formdata['sign'] = self.sign

        res = requests.post(self.url, data=self.formdata, headers=self.head)
        pydict = res.json()
        try:
            dst = jsonpath.jsonpath(pydict, '$..text')[0]      # 单词
        except:
            dst = jsonpath.jsonpath(pydict, '$..dst')[0]       # 句子

        return dst


if __name__ == '__main__':
    zhtoen = BaiduFanyi()
    n = 10
    for i in range(n):
        zh = input("百度翻译为您服务！请输入中文：\n")
        en = zhtoen.fanyi(zh)
        print(f"英文翻译为：\n{en}")
        if i < n - 1:
            y = input(f"你还有{n - i}次机会，继续吗？(Y/N)\n")
            if y in ['n', 'N']:
                break
        else:
            print("机会已经用完，请重新启动...")
            input("请按任意键...")

    print("\n感谢您的使用！")
    time.sleep(1.5)
    print("\n本次服务到此结束！")

*.js文件

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);
    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(""))
    }

    u = "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)
}