某拍房数据采集

声明

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1.逆向目标

点击翻页发包发现sign参数加密

2.寻找加密位置

通过全局搜索关键字发现sign参数特别多，很难找到我们需要的加密位置，可以通过在发包中的发起程序中找到所需要的js文件再进行搜索

进入js文件搜索后，下断点调试，阿里系加密token也是关键词

3.分析加密参数

根据控制台分析

o.token + "&" + a + "&" + s + "&" + n.data

o.token可以直接在请求时的cookie里面获取_m_h5_tk

a的值应该为时间戳，s的值应该为固定参数

n.data为表单数据中的值

通过扣代码将加密代码拿出来，js代码如下

p = function(e) {
        function t(e, t) {
            return e << t | e >>> 32 - t
        }
        function n(e, t) {
            var n, o, r, i, s;
            return r = 2147483648 & e,
            i = 2147483648 & t,
            s = (1073741823 & e) + (1073741823 & t),
            (n = 1073741824 & e) & (o = 1073741824 & t) ? 2147483648 ^ s ^ r ^ i : n | o ? 1073741824 & s ? 3221225472 ^ s ^ r ^ i : 1073741824 ^ s ^ r ^ i : s ^ r ^ i
        }
        function o(e, o, r, i, s, a, p) {
            return e = n(e, n(n(function(e, t, n) {
                return e & t | ~e & n
            }(o, r, i), s), p)),
            n(t(e, a), o)
        }
        function r(e, o, r, i, s, a, p) {
            return e = n(e, n(n(function(e, t, n) {
                return e & n | t & ~n
            }(o, r, i), s), p)),
            n(t(e, a), o)
        }
        function i(e, o, r, i, s, a, p) {
            return e = n(e, n(n(function(e, t, n) {
                return e ^ t ^ n
            }(o, r, i), s), p)),
            n(t(e, a), o)
        }
        function s(e, o, r, i, s, a, p) {
            return e = n(e, n(n(function(e, t, n) {
                return t ^ (e | ~n)
            }(o, r, i), s), p)),
            n(t(e, a), o)
        }
        function a(e) {
            var t, n = "", o = "";
            for (t = 0; 3 >= t; t++)
                n += (o = "0" + (e >>> 8 * t & 255).toString(16)).substr(o.length - 2, 2);
            return n
        }
        var p, u, c, d, l, f, m, g, y, v;
        for (v = function(e) {
            for (var t, n = e.length, o = n + 8, r = 16 * ((o - o % 64) / 64 + 1), i = new Array(r - 1), s = 0, a = 0; n > a; )
                s = a % 4 * 8,
                i[t = (a - a % 4) / 4] = i[t] | e.charCodeAt(a) << s,
                a++;
            return s = a % 4 * 8,
            i[t = (a - a % 4) / 4] = i[t] | 128 << s,
            i[r - 2] = n << 3,
            i[r - 1] = n >>> 29,
            i
        }(e = function(e) {
            e = e.replace(/\r\n/g, "\n");
            for (var t = "", n = 0; n < e.length; n++) {
                var o = e.charCodeAt(n);
                128 > o ? t += String.fromCharCode(o) : o > 127 && 2048 > o ? (t += String.fromCharCode(o >> 6 | 192),
                t += String.fromCharCode(63 & o | 128)) : (t += String.fromCharCode(o >> 12 | 224),
                t += String.fromCharCode(o >> 6 & 63 | 128),
                t += String.fromCharCode(63 & o | 128))
            }
            return t
        }(e)),
        f = 1732584193,
        m = 4023233417,
        g = 2562383102,
        y = 271733878,
        p = 0; p < v.length; p += 16)
            u = f,
            c = m,
            d = g,
            l = y,
            f = o(f, m, g, y, v[p + 0], 7, 3614090360),
            y = o(y, f, m, g, v[p + 1], 12, 3905402710),
            g = o(g, y, f, m, v[p + 2], 17, 606105819),
            m = o(m, g, y, f, v[p + 3], 22, 3250441966),
            f = o(f, m, g, y, v[p + 4], 7, 4118548399),
            y = o(y, f, m, g, v[p + 5], 12, 1200080426),
            g = o(g, y, f, m, v[p + 6], 17, 2821735955),
            m = o(m, g, y, f, v[p + 7], 22, 4249261313),
            f = o(f, m, g, y, v[p + 8], 7, 1770035416),
            y = o(y, f, m, g, v[p + 9], 12, 2336552879),
            g = o(g, y, f, m, v[p + 10], 17, 4294925233),
            m = o(m, g, y, f, v[p + 11], 22, 2304563134),
            f = o(f, m, g, y, v[p + 12], 7, 1804603682),
            y = o(y, f, m, g, v[p + 13], 12, 4254626195),
            g = o(g, y, f, m, v[p + 14], 17, 2792965006),
            f = r(f, m = o(m, g, y, f, v[p + 15], 22, 1236535329), g, y, v[p + 1], 5, 4129170786),
            y = r(y, f, m, g, v[p + 6], 9, 3225465664),
            g = r(g, y, f, m, v[p + 11], 14, 643717713),
            m = r(m, g, y, f, v[p + 0], 20, 3921069994),
            f = r(f, m, g, y, v[p + 5], 5, 3593408605),
            y = r(y, f, m, g, v[p + 10], 9, 38016083),
            g = r(g, y, f, m, v[p + 15], 14, 3634488961),
            m = r(m, g, y, f, v[p + 4], 20, 3889429448),
            f = r(f, m, g, y, v[p + 9], 5, 568446438),
            y = r(y, f, m, g, v[p + 14], 9, 3275163606),
            g = r(g, y, f, m, v[p + 3], 14, 4107603335),
            m = r(m, g, y, f, v[p + 8], 20, 1163531501),
            f = r(f, m, g, y, v[p + 13], 5, 2850285829),
            y = r(y, f, m, g, v[p + 2], 9, 4243563512),
            g = r(g, y, f, m, v[p + 7], 14, 1735328473),
            f = i(f, m = r(m, g, y, f, v[p + 12], 20, 2368359562), g, y, v[p + 5], 4, 4294588738),
            y = i(y, f, m, g, v[p + 8], 11, 2272392833),
            g = i(g, y, f, m, v[p + 11], 16, 1839030562),
            m = i(m, g, y, f, v[p + 14], 23, 4259657740),
            f = i(f, m, g, y, v[p + 1], 4, 2763975236),
            y = i(y, f, m, g, v[p + 4], 11, 1272893353),
            g = i(g, y, f, m, v[p + 7], 16, 4139469664),
            m = i(m, g, y, f, v[p + 10], 23, 3200236656),
            f = i(f, m, g, y, v[p + 13], 4, 681279174),
            y = i(y, f, m, g, v[p + 0], 11, 3936430074),
            g = i(g, y, f, m, v[p + 3], 16, 3572445317),
            m = i(m, g, y, f, v[p + 6], 23, 76029189),
            f = i(f, m, g, y, v[p + 9], 4, 3654602809),
            y = i(y, f, m, g, v[p + 12], 11, 3873151461),
            g = i(g, y, f, m, v[p + 15], 16, 530742520),
            f = s(f, m = i(m, g, y, f, v[p + 2], 23, 3299628645), g, y, v[p + 0], 6, 4096336452),
            y = s(y, f, m, g, v[p + 7], 10, 1126891415),
            g = s(g, y, f, m, v[p + 14], 15, 2878612391),
            m = s(m, g, y, f, v[p + 5], 21, 4237533241),
            f = s(f, m, g, y, v[p + 12], 6, 1700485571),
            y = s(y, f, m, g, v[p + 3], 10, 2399980690),
            g = s(g, y, f, m, v[p + 10], 15, 4293915773),
            m = s(m, g, y, f, v[p + 1], 21, 2240044497),
            f = s(f, m, g, y, v[p + 8], 6, 1873313359),
            y = s(y, f, m, g, v[p + 15], 10, 4264355552),
            g = s(g, y, f, m, v[p + 6], 15, 2734768916),
            m = s(m, g, y, f, v[p + 13], 21, 1309151649),
            f = s(f, m, g, y, v[p + 4], 6, 4149444226),
            y = s(y, f, m, g, v[p + 11], 10, 3174756917),
            g = s(g, y, f, m, v[p + 2], 15, 718787259),
            m = s(m, g, y, f, v[p + 9], 21, 3951481745),
            f = n(f, u),
            m = n(m, c),
            g = n(g, d),
            y = n(y, l);
        return (a(f) + a(m) + a(g) + a(y)).toLowerCase()
    }
console.log(p('1'))

测试后其实就是标准的md5加密

4.python代码书写

大致流程为：

1.先请求主页获取对应cookie值

2.将o.token + "&" + a + "&" + s + "&" + n.data中所需要的参数拼接传入到我们写好的js代码中封装成sign参数再发送请求即可。

下面是关键代码，如需完整代码请私信

获取结果如下