web qq 获取好友列表hash算法

web qq 获取好友列表hash算法

在使用web qq的接口进行好友列表获取的时候，需要post一个参数：hash

在对其js文件进行分析之后，发现计算hash的函数位于：

http://0.web.qstatic.com/webqqpic/pubapps/0/50/eqq.all.js

这个文件中：

		P = function(b, i) {
				for (var a = [], s = 0; s < i.length; s++) a[s % 4] ^= i.charCodeAt(s);
				var j = ["EC", "OK"],
				d = [];
				d[0] = b >> 24 & 255 ^ j[0].charCodeAt(0);
				d[1] = b >> 16 & 255 ^ j[0].charCodeAt(1);
				d[2] = b >> 8 & 255 ^ j[1].charCodeAt(0);
				d[3] = b & 255 ^ j[1].charCodeAt(1);
				j = [];
				for (s = 0; s < 8; s++) j[s] = s % 2 == 0 ? a[s >> 1] : d[s >> 1];
				a = ["0", "1", "2", "3", "4", "5", "6", "7", "8", "9", "A", "B", "C", "D", "E", "F"];
				d = "";
				for (s = 0; s < j.length; s++) d += a[j[s] >> 4 & 15],
				d += a[j[s] & 15];
				return d
		}

这样可以写一个python版本：

		a=[0,0,0,0]
		s=0
		for s in range(0,len(i)):
			a[s%4] = a[s%4] ^ ord(i[s])
		j = ["EC", "OK"]
		d = [0,0,0,0]
		d[0] = int(b) >> 24 & 255 ^ ord(j[0][0])
		d[1] = int(b) >> 16 & 255 ^ ord(j[0][1])
		d[2] = int(b) >> 8 & 255 ^ ord(j[1][0])
		d[3] = int(b) & 255 ^ ord(j[1][1])
		j = range(0,8)
		for s in range(0,8):
			if s % 2 == 0:
				j[s] = a[s >> 1]
			else:
				j[s] = d[s >> 1]
		a = ["0", "1", "2", "3", "4", "5", "6", "7", "8", "9", "A", "B", "C", "D", "E", "F"]
		d = ""
		for s in range(0,len(j)):
			d = d + a[j[s] >> 4 & 15]
			d = d + a[j[s] & 15]
		return d

但是第二天使用web qq接口获取好友列表时，却不能获取了。后来发现原来是这个hash函数变了：

		P = function(b, i) {
			for (var a = [], s = 0; s < b.length; s++) a[s] = b.charAt(s) - 0;  alert(a);//ago
			for (var j = 0, d = -1, s = 0; s < a.length; s++) {
				j += a[s];
				j %=
				i.length;
				var c = 0;
				if (j + 4 > i.length) for (var l = 4 + j - i.length, x = 0; x < 4; x++) c |= x < l ? (i.charCodeAt(j + x) & 255) << (3 - x) * 8: (i.charCodeAt(x - l) & 255) << (3 - x) * 8;
				else for (x = 0; x < 4; x++) c |= (i.charCodeAt(j + x) & 255) << (3 - x) * 8;
				d ^= c
			}
			a = [];
			a[0] = d >> 24 & 255;
			a[1] = d >> 16 & 255;
			a[2] = d >> 8 & 255;
			a[3] = d & 255;
			d = ["0", "1", "2", "3", "4", "5", "6", "7", "8", "9", "A", "B", "C", "D", "E", "F"];
			s = "";
			for (j = 0; j < a.length; j++) s += d[a[j] >> 4 & 15],
			s += d[a[j] & 15];
			return s
		}

继续改写成python版：

		a=[]
		s=0
		for s in range(0,len(b)):
			t=int(b[s])
			a.append(t)
		j = 0
		d = -1
		s = 0
		for s in range(0,len(a)):
			j = j + a[s]
			j = j % len(i)
			c = 0
			if (j + 4) > len(i):
				l = 4 + j - len(i)
				for x in range(0,4):
					if x < l:
						c = c | (( ord(i[j + x]) & 255) << (3 - x) * 8 )
					else:
						c = c | ( ( ord(i[x - l]) & 255) << (3 - x) * 8 )
			else:
				for x in range(0,4):
					c = c | (( ord(i[j + x]) & 255) << (3 - x) * 8 )
			d = d ^ c	

		a = [0,0,0,0]
		a[0] = d >> 24 & 255
		a[1] = d >> 16 & 255
		a[2] = d >> 8 & 255
		a[3] = d & 255
		d = ["0", "1", "2", "3", "4", "5", "6", "7", "8", "9", "A", "B", "C", "D", "E", "F"]
		s = ""
		for j in range(0,len(a)):
			s = s + d[a[j] >> 4 & 15]
			s = s + d[a[j] & 15]
		return s