红联Linux门户
Linux帮助

linux解压.zip文件乱码处理的一个脚本

发布时间:2016-12-26 09:27:20来源:ubuntukylin作者:笨鸟弟弟
使用linux系统往往会下载windows平台编码的文件一般是.zip文件/
 
#!/usr/bin/env python2
# -*- coding: utf-8 -*-
import os
import sys
import zipfile
import argparse
s = '\x1b[%d;%dm%s\x1b[0m'       # terminual color template
def unzip(path):
file = zipfile.ZipFile(path,"r")
if args.secret:
file.setpassword(args.secret)
for name in file.namelist():
try:
utf8name=name.decode('gbk')
pathname = os.path.dirname(utf8name)
except:
utf8name=name
pathname = os.path.dirname(utf8name)
#print s % (1, 92, '  >> extracting:'), utf8name
#pathname = os.path.dirname(utf8name)
if not os.path.exists(pathname) and pathname != "":
os.makedirs(pathname)
data = file.read(name)
if not os.path.exists(utf8name):
try:
fo = open(utf8name, "w")
fo.write(data)
fo.close
except:
pass
file.close()
def main(argv):
######################################################
# for argparse
p = argparse.ArgumentParser(description='解决unzip乱码')
p.add_argument('xxx', type=str, nargs='*', \
help='命令对象.')
p.add_argument('-s', '--secret', action='store', \
default=None, help='密码')
global args
args = p.parse_args(argv[1:])
xxx = args.xxx
for path in xxx:
if path.endswith('.zip'):
if os.path.exists(path):
print s % (1, 97, '  ++ unzip:'), path
unzip(path)
else:
print s % (1, 91, '  !! file doesn\'t exist.'), path
else:
print s % (1, 91, '  !! file isn\'t a zip file.'), path
if __name__ == '__main__':
argv = sys.argv
main(argv)
 
复制代码可以另命名但后缀是py不要忘记他需要python解析。
 
用法:
python2 unzip.py azipfile1.zip azipfile2.zip
python2 unzip.py azipfile.zip -s secret
# -s 密码
 
本文永久更新地址:http://www.linuxdiyf.com/linux/27253.html