unwxapkg.py

#!/usr/bin/env python2
 
# lrdcq 
# usage python2 unwxapkg.py filename
 
import sys, os
import struct
 
class WxapkgFile(object):
  nameLen = 0
  name = ""
  offset = 0
  size = 0


if len(sys.argv) < 2:
  print 'usage: unwxapkg.py filename'
  exit()


with open(sys.argv[1], "rb") as f:
 
  root = os.path.dirname(os.path.realpath(f.name))
  name = os.path.basename(f.name) + '_dir'

  if len(sys.argv) > 2:
    name = sys.argv[2]
 
  #read header
 
  firstMark = struct.unpack('B', f.read(1))[0]
  print 'first header mark = ' + str(firstMark)
 
  info1 = struct.unpack('>L', f.read(4))[0]
  print 'info1 = ' + str(info1)
 
  indexInfoLength = struct.unpack('>L', f.read(4))[0]
  print 'indexInfoLength = ' + str(indexInfoLength)
 
  bodyInfoLength = struct.unpack('>L', f.read(4))[0]
  print 'bodyInfoLength = ' + str(bodyInfoLength)
 
  lastMark = struct.unpack('B', f.read(1))[0]
  print 'last header mark = ' + str(lastMark)
 
  if firstMark != 0xBE or lastMark != 0xED:
    print 'its not a wxapkg file!!!!!'
    exit()
 
  fileCount = struct.unpack('>L', f.read(4))[0]
  print 'fileCount = ' + str(fileCount)
 
  #read index
 
  fileList = []
 
  for i in range(fileCount):
 
    data = WxapkgFile()
    data.nameLen = struct.unpack('>L', f.read(4))[0]
    data.name = f.read(data.nameLen)
    data.offset = struct.unpack('>L', f.read(4))[0]
    data.size = struct.unpack('>L', f.read(4))[0]
 
    print 'readFile = ' + data.name + ' at Offset = ' + str(data.offset)
 
    fileList.append(data)
 
  #save files
   
  for d in fileList:
    d.name = '/' + name + d.name
    path = root + os.path.dirname(d.name)
 
    if not os.path.exists(path):
      os.makedirs(path)
 
    w = open(root + d.name, 'w')
    f.seek(d.offset)
    w.write(f.read(d.size))
    w.close()
 
    print 'writeFile = ' + root + d.name
 
  f.close()

原始代码链接： http://lrdcq.com/me/read.php/66.htm

Amazing tool, Great job!

You can find a python3 version here

解出来是 一些 .html 文件，编辑器都打不开。。

请问怎么获取小程序的appID？

解压出来的都是html文件，怎么用呢

同问 HTML文件有还原的工具吗

Traceback (most recent call last):
File "unwxapkg.py", line 78, in
w = open(root + d.name, 'w')
IOError: [Errno 22] invalid mode ('w') or filename: 'xxx.wxapkg_dir/res/\xe6\xb0\x94\xe6\xb3\xa1\xe7\x8b\x97\xe7\x99\xbd\xe8\x84\xb8.png'

Traceback (most recent call last):
File "unwxapkg.py", line 31, in
firstMark = struct.unpack('B', f.read(1))[0]
struct.error: unpack requires a string argument of length 1
这个是啥毛病....

同问html后面怎么解码

可参考 https://github.com/chenrensong/SS.UnWxapkg

https://wxbug.cn 这里可直接上传解析

现在微信小游戏里的图片加密了？能破吗

如果出现中文文件名会报错，可以试试这样（ line: 65 && line: 78）：

# line 65
print 'readFile = ' + data.name.decode('utf-8') + ' at Offset = ' + str(data.offset)

# line 78
filepath = unicode(root + d.name,'utf8')
w = open(filepath, 'w')