應(yīng)用處景

如不雅須要批量分析apk以及每個(gè)apk文件中的classes.dex 文件。怎么提掏出它們？將apk改后綴名變?yōu)?zip文件，之后在解壓，提掏出每個(gè)apk文件中的classes.dex文件，這是一個(gè)可行的籌劃。然則中心解壓大年夜量的apk文件會(huì)占據(jù)我們的大年夜量磁盤存儲(chǔ)空間，怎么在不解壓文件的情況下提掏出dex文件？在這里應(yīng)用python自帶的zipfile類，可以輕松的解決這個(gè)問題。

代碼實(shí)現(xiàn)：

#!/usr/bin/env python
# coding=utf-8

'''
@author   : Chicho
@version  : 1.0
@ date    : Jan 4, 2015 01:54
@function : extract dexs file from apk files
               * create a new directory to store all dex files

@running  : python dex_extract.py
'''


import os
import zipfile


path="/home/chicho/test/test/"  # this is apk files' store path
dex_path="/home/chicho/test/test/dex/" # a directory  store dex files 


apklist = os.listdir(path) # get all the names of apps

if not os.path.exists(dex_path):
    os.makedirs(dex_path)

for APK in apklist:
    portion = os.path.splitext(APK)

    if portion[1] == ".apk":
        newname = portion[0] + ".zip" # change them into zip file to extract dex files

        os.rename(APK,newname)

    if APK.endswith(".zip"):
        apkname = portion[0]

        zip_apk_path = os.path.join(path,APK) # get the zip files

        z = zipfile.ZipFile(zip_apk_path, 'r') # read zip files

        for filename in z.namelist():
            if filename.endswith(".dex"):
                dexfilename = apkname + ".dex"
                dexfilepath = os.path.join(dex_path, dexfilename)
                f = open(dexfilepath, 'w+') # eq: cp classes.dex dexfilepath
                f.write(z.read(filename))



print "all work done!"

應(yīng)用這個(gè)辦法還可以大年夜apk文件中批量的提掏出其他文件。比如說so文件等。
轉(zhuǎn)載注明出處：http://blog.csdn.net/chichoxian/article/details/42382695