微信饿了么外卖返利电影票小程序公众号分销商城定制设计H5源码

本文分享了一款外卖CPS红包小程序的源码,包括饿了么和美团优惠券功能,具备分销返利特性。提供了源代码下载地址、搭建步骤和操作指南,适用于微信小程序开发。

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外卖CPS红包小程序源码分享

外卖cps带分销返利源码

饿了么红包、美团优惠券开发(外卖cps,三级裂变源码)

源代码地址

http://y.mybei.cn

搭建步骤

截图

在这里插入图片描述

步骤

(wx.webpackJsonp = wx.webpackJsonp || []).push([ [ 23 ], {
176: function(r, t, e) {
r.exports = e.p + “npm/taro-code/lib/weapp/components/QRCode/index.wxml”;
},
432: function(r, t, e) {
“use strict”;
e(176);
},
437: function(r, t, e) {
“use strict”;
e.r(t);
e(432);
var n = e(0), o = e.n(n), a = e(1), i = e.n(a), u = function qrcode(r, t) {
var e = r, n = v[t], o = null, a = 0, i = null, u = new Array(), f = {}, c = function makeImpl(r, t) {
o = function® {
for (var t = new Array®, e = 0; e < r; e += 1) {
t[e] = new Array®;
for (var n = 0; n < r; n += 1) t[e][n] = null;
}
return t;
}(a = 4 * e + 17), s(0, 0), s(a - 7, 0), s(0, a - 7), g(), l(), p(r, t), e >= 7 && h®,
null == i && (i = w(e, n, u)), d(i, t);
}, s = function setupPositionProbePattern(r, t) {
for (var e = -1; e <= 7; e += 1) if (!(r + e <= -1 || a <= r + e)) for (var n = -1; n <= 7; n += 1) t + n <= -1 || a <= t + n || (o[r + e][t + n] = 0 <= e && e <= 6 && (0 == n || 6 == n) || 0 <= n && n <= 6 && (0 == e || 6 == e) || 2 <= e && e <= 4 && 2 <= n && n <= 4);
}, l = function setupTimingPattern() {
for (var r = 8; r < a - 8; r += 1) null == o[r][6] && (o[r][6] = r % 2 == 0);
for (var t = 8; t < a - 8; t += 1) null == o[6][t] && (o[6][t] = t % 2 == 0);
}, g = function setupPositionAdjustPattern() {
for (var r = A.getPatternPosition(e), t = 0; t < r.length; t += 1) for (var n = 0; n < r.length; n += 1) {
var a = r[t], i = r[n];
if (null == o[a][i]) for (var u = -2; u <= 2; u += 1) for (var f = -2; f <= 2; f += 1) o[a + u][i + f] = -2 == u || 2 == u || -2 == f || 2 == f || 0 == u && 0 == f;
}
}, h = function setupTypeNumber® {
for (var t = A.getBCHTypeNumber(e), n = 0; n < 18; n += 1) {
var i = !r && 1 == (t >> n & 1);
o[Math.floor(n / 3)][n % 3 + a - 8 - 3] = i;
}
for (n = 0; n < 18; n += 1) {
i = !r && 1 == (t >> n & 1);
o[n % 3 + a - 8 - 3][Math.floor(n / 3)] = i;
}
}, p = function setupTypeInfo(r, t) {
for (var e = n << 3 | t, i = A.getBCHTypeInfo(e), u = 0; u < 15; u += 1) {
var f = !r && 1 == (i >> u & 1);
u < 6 ? o[u][8] = f : u < 8 ? o[u + 1][8] = f : o[a - 15 + u][8] = f;
}
for (u = 0; u < 15; u += 1) {
f = !r && 1 == (i >> u & 1);
u < 8 ? o[8][a - u - 1] = f : u < 9 ? o[8][15 - u - 1 + 1] = f : o[8][15 - u - 1] = f;
}
o[a - 8][8] = !r;
}, d = function mapData(r, t) {
for (var e = -1, n = a - 1, i = 7, u = 0, f = A.getMaskFunction(t), c = a - 1; c > 0; c -= 2) for (6 == c && (c -= 1); 😉 {
for (var s = 0; s < 2; s += 1) if (null == o[n][c - s]) {
var l = !1;
u < r.length && (l = 1 == (r[u] >>> i & 1)), f(n, c - s) && (l = !l), o[n][c - s] = l,
-1 == (i -= 1) && (u += 1, i = 7);
}
if ((n += e) < 0 || a <= n) {
n -= e, e = -e;
break;
}
}
}, w = function createData(r, t, e) {
for (var n = x.getRSBlocks(r, t), o = D(), a = 0; a < e.length; a += 1) {
var i = e[a];
o.put(i.getMode(), 4), o.put(i.getLength(), A.getLengthInBits(i.getMode(), r)),
i.write(o);
}
var u = 0;
for (a = 0; a < n.length; a += 1) u += n[a].dataCount;
if (o.getLengthInBits() > 8 * u) throw new Error(“code length overflow. (” + o.getLengthInBits() + “>” + 8 * u + “)”);
for (o.getLengthInBits() + 4 <= 8 * u && o.put(0, 4); o.getLengthInBits() % 8 != 0; ) o.putBit(!1);
for (;!(o.getLengthInBits() >= 8 * u || (o.put(236, 8), o.getLengthInBits() >= 8 * u)); ) o.put(17, 8);
return function createBytes(r, t) {
for (var e = 0, n = 0, o = 0, a = new Array(t.length), i = new Array(t.length), u = 0; u < t.length; u += 1) {
var f = t[u].dataCount, c = t[u].totalCount - f;
n = Math.max(n, f), o = Math.max(o, c), a[u] = new Array(f);
for (var s = 0; s < a[u].length; s += 1) a[u][s] = 255 & r.getBuffer()[s + e];
e += f;
var l = A.getErrorCorrectPolynomial©, g = qrPolynomial(a[u], l.getLength() - 1).mod(l);
i[u] = new Array(l.getLength() - 1);
for (s = 0; s < i[u].length; s += 1) {
var h = s + g.getLength() - i[u].length;
i[u][s] = h >= 0 ? g.getAt(h) : 0;
}
}
var p = 0;
for (s = 0; s < t.length; s += 1) p += t[s].totalCount;
var v = new Array§, d = 0;
for (s = 0; s < n; s += 1) for (u = 0; u < t.length; u += 1) s < a[u].length && (v[d] = a[u][s],
d += 1);
for (s = 0; s < o; s += 1) for (u = 0; u <

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