五矿webpack流程分析二进行逆向

最新推荐文章于 2025-12-18 09:44:24 发布

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最新推荐文章于 2025-12-18 09:44:24 发布 · 419 阅读

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#webpack #前端 #node.js

加密的地方是t.encryptLong

往上看t的生成位置t = new d["a"]

d = t("9816");

我们往上找入口文件

往下看 webpackmod

由于我们找到了加密位置可以直接把加密位置那块扣出来

function rsa(key,page) {
    r = key
    t.setPublicKey(r)
    e = {"inviteMethod": "", "businessClassfication": "", "mc": "", "lx": "ZBGG", "dwmc": "", "pageIndex": page}
    a = m(m({}, e), {}, {sign: md5(JSON.stringify(e)), timeStamp: +new Date})
    console.log(a)
    s = t.encryptLong(JSON.stringify(a))
    return s
}

入口

(
    function (A) {
        function e(e) {
            for (var n, r, i = e[0], o = e[1], c = e[2], g = 0, C = []; g < i.length; g++)
                r = i[g],
                Object.prototype.hasOwnProperty.call(a, r) && a[r] && C.push(a[r][0]),
                    a[r] = 0;
            for (n in o)
                Object.prototype.hasOwnProperty.call(o, n) && (A[n] = o[n]);
            l && l(e);
            while (C.length)
                C.shift()();
            return s.push.apply(s, c || []),
                t()
        }

        function t() {
            for (var A, e = 0; e < s.length; e++) {
                for (var t = s[e], n = !0, r = 1; r < t.length; r++) {
                    var i = t[r];
                    0 !== a[i] && (n = !1)
                }
                n && (s.splice(e--, 1),
                    A = o(o.s = t[0]))
            }
            return A
        }

        var n = {}
            , r = {
            index: 0
        }
            , a = {
            index: 0
        }
            , s = [];

        function i(A) {
            return o.p + "js/" + ({}[A] || A) + ".js"
        }

        function o(e) {
            if (n[e])
                return n[e].exports;
            var t = n[e] = {
                i: e,
                l: !1,
                exports: {}
            };
            console.log(e)
            return A[e].call(t.ex