python - 将连接值传递给python中的zipfile方法

Question

我是python的初学者。我在将函数参数与字符串连接并将其传递给 Zipfile 时遇到问题。请帮我解决这个问题，

import zipfile,os
from xml.dom import minidom
def zipFunc(output)
absPath='E:\'+output+'.zip'
fName='package'
with zipfile.ZipFile(absPath, 'r') as zip:
data=zip.read(fName+'/manifest.txt')
xmldoc =minidom.parseString(data)
itemlist = xmldoc.getElementsByTagName('value') 
for s in itemlist :
        if s.attributes['name'].value=="version":
            print s.firstChild.nodeValue

最终的连接路径应该是，

E:\pyauto\package.zip

但我面临以下错误，

with zipfile.ZipFile(absPath, 'r') as zip:
File "C:\Python27\lib\zipfile.py", line 756, in __init__
self.fp = open(file, modeDict[mode])
IOError: [Errno 22] invalid mode ('rb') or filename: 
u'E:\\pyauto\\package\r\x07.zip'

标签： pythonconcatenationzipfile