Python爬虫 js逆向-----百度翻译

最新推荐文章于 2023-11-18 13:17:32 发布

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最新推荐文章于 2023-11-18 13:17:32 发布

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文章标签：爬虫 python

本文链接：https://blog.youkuaiyun.com/big_god_/article/details/120861842

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js逆向之百度翻译

声明：本篇博客涉及到的网址只做学习使用，请勿做其他用途

本小菜鸟最近在学习爬虫js逆向，这是爬虫进阶的第一步，刚好找到了百度翻译来做了一个日常训练，这个网站刚好适合准备进阶的小白学习，好了，话不多说，咱们直接进入正题。
第一步：在翻译框内输入文字，并打开F12
在打开调试窗口的时候跳过debug断点抓包，本菜鸟测试发现呢，中文翻译的post请求与Form Data里面的sign有关系，每个翻译的中文所对应的sign的值是不同的，所以我们必须拿到这个参数
在这里插入图片描述
第二步：通过我们全局搜索sign这个参数，我们发现sign是通过L(e)这个js函数生成的

在这里插入图片描述
所以我们打个断点进入到这个函数里面

我们发现这个函数不能用Python来还原，所以直接将这段js代码扣下来在nodejs里面运行，函数function e（r）传入的的r值就是要翻译的文字,然后通过Python调用这段js代码

//baidu.js  给e函数重新命名get_sign 方便后面调用
function get_sign(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(""))
        }
        var u = void 0
          , l = "" + String.fromCharCode(103) + String.fromCharCode(116) + String.fromCharCode(107);
        u = null !== i ? i : (i = window[l] || "") || "";
        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)
    }

import execjs
def get_sign(word):
    with open('baidu.js','r',encoding='utf8')as f:
        jscode = f.read()
    sign = execjs.compile(jscode).call('get_sign',word)
    return sign
get_sign('你好')

发现报错了，并且提示我们缺少参数i
在这里插入图片描述
所以我们就要补齐这些缺少的值，将鼠标移动到 i 这个值上面，发现i为一个固定的值，所以我们直接在js里面定义

然后测试，发现又缺少n和a,我们依旧补齐，补齐后的js代码为：


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 a(r) {
    if (Array.isArray(r)) {
        for (var o = 0, t = Array(r.length); o < r.length; o++)
            t[o] = r[o];
        return t
    }
    return Array.from(r)
    }

function get_sign(r) {
    var i = '320305.131321201'
    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(""))
    }
    var u = void 0
      , l = "" + String.fromCharCode(103) + String.fromCharCode(116) + String.fromCharCode(107);
    u = null !== i ? i : (i = window[l] || "") || "";
    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)
    }