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