qq好友列表获取hash算法解析-web版QQ
在使用web qq的接口进行qq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接口实现qq好友列表获取时，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更多qq好友列表获取实现方法 可以加Q：2163927521 一起交流