ARMILKCHEN
V2EX  ›  Node.js

爬虫遇到 js 加密问题 想请教一下前端大哥们

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  •   ARMILKCHEN · Apr 29, 2020 · 6746 views
    This topic created in 2379 days ago, the information mentioned may be changed or developed.
    不懂为什么这个 JSON.parse 可以将数据修改了?我应该怎么进行逆向去找?
    13 replies  •  2020-04-30 09:45:15 +08:00
    ARMILKCHEN
    ARMILKCHEN
    OP
    Apr 29, 2020
    1
    s 是正确的数据 l 是进行加密后的数据
    noreplay
    noreplay
    Apr 29, 2020 via Android
    2
    啥都看不出来…
    ARMILKCHEN
    ARMILKCHEN
    OP
    Apr 29, 2020
    3
    Delon
    Delon
    Apr 29, 2020
    4
    parse 被 hook 掉了
    ARMILKCHEN
    ARMILKCHEN
    OP
    Apr 29, 2020
    5
    @Delon #4 那应该怎么解决 可以提供下思路吗
    1018ji
    1018ji
    Apr 29, 2020
    7
    @ARMILKCHEN 看链接,剩下的自己搞吧
    ARMILKCHEN
    ARMILKCHEN
    OP
    Apr 29, 2020
    8
    @1018ji #7 多谢!方便告诉我是怎么找到 e1(e2(str))的执行吗?
    crab
    crab
    Apr 29, 2020
    9
    u.c(u.d(JSON.stringify(l.payload)))

    u.d=e1
    u.c=e2


    function e2(e) {
    if (null == (e = _u_e(e)))
    return null;
    for (var t = "", n = 0; n < e.length; n++) {
    var r = _p.charCodeAt(n % _p.length);
    t += String.fromCharCode(e.charCodeAt(n) ^ r)
    }
    return t

    function e1(e) {
    if (null == e)
    return null;
    for (var t, n, r, o, i, a, u, c = "", s = 0; s < e.length; )
    o = (t = e.charCodeAt(s++)) >> 2,
    i = (3 & t) << 4 | (n = e.charCodeAt(s++)) >> 4,
    a = (15 & n) << 2 | (r = e.charCodeAt(s++)) >> 6,
    u = 63 & r,
    isNaN(n) ? a = u = 64 : isNaN(r) && (u = 64),
    c = c + _keyStr.charAt(o) + _keyStr.charAt(i) + _keyStr.charAt(a) + _keyStr.charAt(u);
    return c
    }
    crab
    crab
    Apr 29, 2020
    10
    @ARMILKCHEN 断下后,F11 单步走就跟到了。
    1018ji
    1018ji
    Apr 29, 2020
    11
    @ARMILKCHEN 楼上大佬说的很清楚了,一步步找就行。我一般是 charles 把 js 整在本地测试打函数
    ARMILKCHEN
    ARMILKCHEN
    OP
    Apr 30, 2020
    12
    @1018ji #11 好的 多谢!
    ARMILKCHEN
    ARMILKCHEN
    OP
    Apr 30, 2020
    13
    @crab #10 谢谢大佬!
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