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
Santiago Trujillo
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)
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.
Try calling zipfile.close() afterwards?
from zipfile import ZipFile
zipf = ZipFile("main.zip","w", zipfile.ZIP_DEFLATED)
zipf.write("main.json")
zipf.close()