Zip single file

Question

I am trying to zip a single file in python. For whatever reason, I'm having a hard time getting down the syntax. What I am trying to do is keep the original file and create a new zipped file of the original (like what a Mac or Windows would do if you archive a file).

Here is what I have so far:

import zipfile

myfilepath = '/tmp/%s' % self.file_name
myzippath = myfilepath.replace('.xml', '.zip')

zipfile.ZipFile(myzippath, 'w').write(open(myfilepath).read()) # does not zip the file properly

Answer 1

If the file to be zipped (filename) is in a different directory called pathname, you should use the arcname parameter. Otherwise, it will recreate the full folder hierarchy to the file folder.

from zipfile import ZipFile
import os

with ZipFile(zip_file, 'w') as zipf:
    zipf.write(os.path.join(pathname,filename), arcname=filename)

Answer 2

The correct way to zip file is:

zipfile.ZipFile('hello.zip', mode='w').write("hello.csv")
# assume your xxx.py under the same dir with hello.csv

The python official doc says:

ZipFile.write(filename, arcname=None, compress_type=None)

Write the file named filename to the archive, giving it the archive name arcname

You pass open(filename).read() into write(). open(filename).read() is a single string that contains the whole content of file filename, it would throw FileNotFoundError because it is trying to find a file named with the string content.

Answer 3

Try calling zipfile.close() afterwards?

   from zipfile import ZipFile
   zipf = ZipFile("main.zip","w", zipfile.ZIP_DEFLATED)
   zipf.write("main.json")

   zipf.close()

Answer 4

Since you also want to specify the directory try using os.chdir:

#!/usr/bin/python

from zipfile import ZipFile
import os

os.chdir('/path/of/target/and/destination')
ZipFile('archive.zip', 'w').write('original_file.txt')

• Python zipfile : Work with Zip archives
• Python Miscellaneous operating system interfaces