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