整体过程和h5st网页版差不多。感觉小程序对环境检测并不是很严格。访问多次太多直接封openid。加密流程差不都。首先对body加密。加密后把整个函数扣下来调用一下就行了。
- sha256_digest=function(r) {
- function n(r, n) {
- var t = (65535 & r) + (65535 & n);
- return (r >> 16) + (n >> 16) + (t >> 16) << 16 | 65535 & t
- }
- function t(r, n) {
- return r >>> n | r << 32 - n
- }
- function e(r, n) {
- return r >>> n
- }
- function o(r, n, t) {
- return r & n ^ ~r & t
- }
- function u(r, n, t) {
- return r & n ^ r & t ^ n & t
- }
- function a(r) {
- return t(r, 2) ^ t(r, 13) ^ t(r, 22)
- }
- function f(r) {
- return t(r, 6) ^ t(r, 11) ^ t(r, 25)
- }
- function i(r) {
- return t(r, 7) ^ t(r, 18) ^ e(r, 3)
- }
- function c(r) {
- return t(r, 17) ^ t(r, 19) ^ e(r, 10)
- }
- var h = 8
- , C = 0;
- return r = function(r) {
- r = r.replace(/\r\n/g, "\n");
- for (var n = "", t = 0; t < r.length; t++) {
- var e = r.charCodeAt(t);
- e < 128 ? n += String.fromCharCode(e) : e > 127 && e < 2048 ? (n += String.fromCharCode(e >> 6 | 192),
- n += String.fromCharCode(63 & e | 128)) : (n += String.fromCharCode(e >> 12 | 224),
- n += String.fromCharCode(e >> 6 & 63 | 128),
- n += String.fromCharCode(63 & e | 128))
- }
- return n
- }(r),
- function(r) {
- for (var n = C ? "0123456789ABCDEF" : "0123456789abcdef", t = "", e = 0; e < 4 * r.length; e++)
- t += n.charAt(r[e >> 2] >> 8 * (3 - e % 4) + 4 & 15) + n.charAt(r[e >> 2] >> 8 * (3 - e % 4) & 15);
- return t
- }(function(r, t) {
- var e, h, C, g, d, v, A, l, m, S, s = new Array(1116352408,1899447441,3049323471,3921009573,961987163,1508970993,2453635748,2870763221,3624381080,310598401,607225278,1426881987,1925078388,2162078206,2614888103,3248222580,3835390401,4022224774,264347078,604807628,770255983,1249150122,1555081692,1996064986,2554220882,2821834349,2952996808,3210313671,3336571891,3584528711,113926993,338241895,666307205,773529912,1294757372,1396182291,1695183700,1986661051,2177026350,2456956037,2730485921,2820302411,3259730800,3345764771,3516065817,3600352804,4094571909,275423344,430227734,506948616,659060556,883997877,958139571,1322822218,1537002063,1747873779,1955562222,2024104815,2227730452,2361852424,2428436474,2756734187,3204031479,3329325298), y = new Array(1779033703,3144134277,1013904242,2773480762,1359893119,2600822924,528734635,1541459225), w = new Array(64);
- r[t >> 5] |= 128 << 24 - t % 32,
- r[15 + (t + 64 >> 9 << 4)] = t;
- for (var p = 0; p < r.length; p += 16) {
- e = y[0],
- h = y[1],
- C = y[2],
- g = y[3],
- d = y[4],
- v = y[5],
- A = y[6],
- l = y[7];
- for (var b = 0; b < 64; b++)
- w[b] = b < 16 ? r[b + p] : n(n(n(c(w[b - 2]), w[b - 7]), i(w[b - 15])), w[b - 16]),
- m = n(n(n(n(l, f(d)), o(d, v, A)), s[b]), w[b]),
- S = n(a(e), u(e, h, C)),
- l = A,
- A = v,
- v = d,
- d = n(g, m),
- g = C,
- C = h,
- h = e,
- e = n(m, S);
- y[0] = n(e, y[0]),
- y[1] = n(h, y[1]),
- y[2] = n(C, y[2]),
- y[3] = n(g, y[3]),
- y[4] = n(d, y[4]),
- y[5] = n(v, y[5]),
- y[6] = n(A, y[6]),
- y[7] = n(l, y[7])
- }
- return y
- }(function(r) {
- for (var n = Array(), t = (1 << h) - 1, e = 0; e < r.length * h; e += h)
- n[e >> 5] |= (r.charCodeAt(e / h) & t) << 24 - e % 32;
- return n
- }(r), r.length * h))
- }
但是需要主要token,和fp。这两个手动调用一下函数赋值给fe就行了。还有一些小细节就不说了。