create a zipfile from an input file - Python Snipplr Social Repository

Revision: 26573

at May 1, 2010 12:28 by throb

Initial Code

import zlib, zipfile, os

def zipme (inputFilename):
    fileName = os.path.basename(inputFilename)
    zipFileName = os.path.basename (os.path.splitext(fileName)[0]) + '.zip'
    zipFilePath = os.path.dirname (inputFilename)
    zFile = zipfile.ZipFile(zipFilePath + '/' + zipFileName ,'w')
    zFile.write (inputFilename, fileName, zipfile.ZIP_DEFLATED)
    zFile.close()
    if os.stat(zipFileName):
        print 'zipfile created : %s' % (zipFileName)
        return 1
    else :
        print 'Error creating zipfile : %s' % (zipFileName)
        return 0

Initial URL

Initial Description

This creates a zipfile with the same name as the input file in the same location as the input file.  Only does a single file.

Initial Title

create a zipfile from an input file

Initial Tags

python

Initial Language

Python

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