2022记某东自动化抢购流程实现（实际没啥卵用）
不知不觉又到年底了，为了让这个年有点仪式感，我觉着少了一瓶茅台，又不想当冤大头，只能去某东去抢，去年的自动化脚本已经完全失效了。只能从头看接口重写一个了。以下内容是完成此目标必须解决的三个主要问题：
第一个问题：sk参数
最直接的想法就是把能提取参数先临时提取出来，直接模拟最后submitOrder就完事了，但是，
仔细一对，发现submit请求的参数中有个每次请求都会变的参数,即 sk，如下图中的 sk: JKMCKRY9R9OIYXKXHDPAPUA2
它是由https://tak.jd.com/t/xxxx?_t=16xxxxxxx65（第二个红框）请求的结果
{
"code": "000000",
"message": "success",
"data": {
"@t": "ch",
"j": "A19JMSnoGQB9O9O44Pr0",
"ibV": "BqytgL31urDQOURzpVbj",
"EUK": "JkMCkRY9r9oIyXkXhDpa",
"JRy": "NbPqD8pUa2cL8qGlQeSJ",
"R": "llUnV7u1d4v3tRmYrHrU"
}
}根据一定的规则生成的，而这个计算规则，则在第一个红框所请求的高度混淆的js（第一个红框）中，如下：
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case '6':
P[bq(0x173) + 'd'](R);
continue;
case '7':
var R = E[bq(0x14e) + bq(0x1e4)](bq(0x164));
continue;
}
break;
}
} else {
function S() {
var br = bq;
return E[br(0x1f9)];
}
}
} catch (T) {
if (c[bq(0x12a)] !== c['yDAax']) return ![];
else {
function U() {
var bs = bq,
V = M['Vvhpk'][bs(0xdf)]('|'),
W = 0x0;
while (!![]) {
switch (V[W++]) {
case '0':
Z['removeChil' + 'd'](X);
continue;
case '1':
var X = X[bs(0x14e) + bs(0x1e4)](M[bs(0x21a)]);
continue;
case '2':
Z['insertBefo' + 're'](X, Z[bs(0x1a8)]);
continue;
case '3':
X[bs(0x19e)] = M['WfeMF'](Y, M[bs(0x245)]);
continue;
case '4':
if (M[bs(0x25a)](l[Y], m[Y])) return;
continue;
case '5':
return n(M[bs(0x127)]);
case '6':
var Y = M[bs(0x1c4)](bs(0x124), M[bs(0xff)](new k(), 0x0));
continue;
case '7':
var Z = X[bs(0x20b) + bs(0x162)]('head')[0x0];
continue;
case '8':
if (M[bs(0x219)](typeof i, M['hljMP'])) return j(M[bs(0xe1)]);
continue;
}
break;
}
}
}
}
}
function H(L) {
var bt = a6;
try {
if (c[bt(0x249)](bt(0x224), c[bt(0x179)])) return JSON[bt(0x1ac)](L);
else {
function M() {
var bu = bt,
N = '';
for (var O = 0x0; O < 0x7; O++) {
var P = c[bu(0xde)](i, j['length']);
N = c[bu(0x121)](N + k['charAt'](P), P);
}
return N;
}
}
} catch (N) {}
return {};
}
function I(L) {
var bv = a6;
if (c['CBUYZ'](c[bv(0x21d)], c[bv(0x253)])) return J(h), L ? L : c['XMngE'](k, c['QZsnf']);
else {
function M() {
var bw = bv;
return e[bw(0x1ac)](f);
}
}
}
function J(L) {
var bx = a6;
if (typeof L !== c[bx(0x1c6)]) return k('618qaz');
var M = c[bx(0x1be)](c[bx(0x10a)], c[bx(0x10c)](new Date(), 0x0)),
N = E['createElem' + 'ent'](c[bx(0x123)]),
O = E[bx(0x20b) + bx(0x162)](c['RCWbV'])[0x0];
O['insertBefo' + 're'](N, O['firstChild']), N['text'] = M + c[bx(0x256)], O['removeChil' + 'd'](N);
if (c[bx(0xfb)](D[M], L[M])) {
if (c[bx(0x203)](c[bx(0x178)], c[bx(0x178)])) {
function P() {
return E;
}
} else return;
}
return c[bx(0xde)](k, c[bx(0xe8)]);
}
var K = function () {
var bB = a6,
L = {
'ioCGU': function (O, P) {
var by = b;
return c[by(0x137)](O, P);
}, 'FhOZt': function (O, P) {
var bz = b;
return c[bz(0x25d)](O, P);
}, 'WTzan': function (O, P) {
return c['RLJEj'](O, P);
}, 'nApJv': function (O, P) {
var bA = b;
return c[bA(0x1db)](O, P);
}, 'kVNZW': bB(0x1fb),
'GJcnm': function (O, P) {
var bC = bB;
return c[bC(0x203)](O, P);
}, 'rsSaL': c[bB(0x120)],
'sgZjN': c[bB(0x205)],
'NJpfY': function (O, P) {
var bD = bB;
return c[bD(0x1b0)](O, P);
}, 'yMvBC': function (O, P) {
return O(P);
}, 'EqacN': function (O, P) {
return c['ZNrPS'](O, P);
}, 'ggGUD': c[bB(0x160)],
'wVqTU': function (O, P) {
return c['zcyjY'](O, P);
}, 'gevUl': function (O, P) {
return O(P);
}
},
M = '';
try {
if (c[bB(0x126)](c[bB(0x194)], c['ERTbH'])) {
function O() {
var bE = bB;
xrRZjk[bE(0x23b)](e[bE(0x13e) + 'e'](), f[bE(0x1a1)](0x6, 0xa)[bE(0x13e) + 'e']());
}
} else {
var N = {};
N[c[bB(0x10b)]] = c[bB(0x1db)](c[bB(0x161)](I(e), bB(0x229)), new Date()['getTime']()), N[bB(0x257)] = ![], N[bB(0xe7)] = 0x12c, N[c[bB(0x1ff)]] = function (P) {
var bH = bB,
Q = {
'dQPfJ': function (T, U) {
var bF = b;
return L[bF(0x141)](T, U);
}, 'qNjYF': function (T, U) {
var bG = b;
return L[bG(0x111)](T, U);
}
};
if (L[bH(0x1dc)] === L[bH(0x1dc)]) try {
if (L[bH(0x1a2)](L[bH(0x23c)], L[bH(0x19c)])) {
P = L[bH(0x1a5)](H, P);
if (L[bH(0x181)](I, P) && I(P['data'])) {
var R = [];
for (var S in P['data']) {
if (L[bH(0x129)](L[bH(0x266)], L[bH(0x266)])) R['push'](L[bH(0xf8)](I, P[bH(0x23a)][S]));
else {
function T() {
d;
}
}
}
M = L[bH(0x134)](I, F[bH(0x153)](this, R));
}
} else {
function U() {
var bI = bH;
g += Q['dQPfJ'](Q['dQPfJ'](Q[bI(0x196)](h, '='), i[j]), '&');
}
}
} catch (V) {} else {
function W() {
var bJ = bH;
return f[bJ(0x1fd)](L[bJ(0x255)](g[bJ(0xe2)](), h));
}
}
}, c[bB(0x1b0)](u, c[bB(0x1b0)](I, N));
}
} catch (P) {}
return !M && (M = c[bB(0x237)](C)), c['zcyjY'](I, M);
};
if (o()) K(), _tak[a6(0x1ea)] = function () {
var bK = a6,
L = {
'GHabM': function (M) {
return c['MUqNz'](M);
}
};
if (c[bK(0x247)] === c[bK(0x223)]) {
function M() {
var bL = bK;
return E[bL(0x208)];
}
} else return g(function () {
var bM = bK;
L[bM(0x1da)](K);
}, 0x64), r ? r : c[bK(0x258)](C);
};
else {
if (c['hGrJU'](a6(0x209), c[a6(0xf2)])) _tak[a6(0x1ea)] = K;
else {
function L() {
var bN = a6;
g && (c[bN(0x1b0)](k, l), m = null);
}
}
}
}();
} catch (a0) {}看到这，差点直接被劝退。硬着头皮一顿debug 替换，总算把这个js逻辑看明白了，sk的生成逻辑算是搞定了，如下：
这里只能花功夫了，没啥捷径了。
第二个问题：流程
完成上面工作我一度以为已经可以了，但是还不行，直接提交会被直接拒绝连接，这是submit请求缺少了关键cookie mid，而这个mid是 https://api.m.jd.com/client.action?functionId=genToken获取跳转tokenKey后请求https://un.m.jd.com/cgi-bin/app/appjmp?tokenKey=xxxxxxx&to=https%3A%2F%2Fdivide.jd.com%2Fuser_routing%3FskuId%3D100012043978%26from%3Dapp完成一系列302跳转时种下的，所以必须要完成这套流程
第三个问题：签名sign
要完成第二个问题，tokenKey的获取，需要进行api请求，而api请求，每次请求需要对请求body进行签名，而这个签名每次也是实时计算的，且和时间戳有关，如 st=16xxxxxxxxxx&sign=cd59ab129b6d43b975318f9f6879e030&sv=121,其中 st是时间戳，这里打码处理下。签名算法是隐藏在APP端so库中的。这方面有很多人做过工作了，我也是借鉴了前人的工作，这里也不再啰嗦了。这部分工作也是只能花功夫的，没啥捷径。
最后
再解决了以上问题后，自动化流程终于可以跑了，效果如下:
目标尚未达成，遥遥无期。。。
结论
对于抢购并没有什么太大帮助，现在应该不是按时间先后了，获取mid的过程其实本质就是摇取抢购资格令牌的过程，一旦有这个机制在，那也只能看运气了
能不能分享一下源码啊，学习一下
你好这个sk参数怎么通过js文件传递啊
大佬，太牛了！
最新代码有么，求分享～