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create a zipfile from an input file

/ Published in: Python
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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.

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  1. import zlib, zipfile, os
  2.  
  3. def zipme (inputFilename):
  4.     fileName = os.path.basename(inputFilename)
  5.     zipFileName = os.path.basename (os.path.splitext(fileName)[0]) + '.zip'
  6.     zipFilePath = os.path.dirname (inputFilename)
  7.     zFile = zipfile.ZipFile(zipFilePath + '/' + zipFileName ,'w')
  8.     zFile.write (inputFilename, fileName, zipfile.ZIP_DEFLATED)
  9.     zFile.close()
  10.     if os.stat(zipFileName):
  11.         print 'zipfile created : %s' % (zipFileName)
  12.         return 1
  13.     else :
  14.         print 'Error creating zipfile : %s' % (zipFileName)
  15.         return 0

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