How to take a zipfile as a File object in django?

Question

I have a folder named 'src' of which I create a zip file named zf. I want to pass zf as a FileField object to newobj.fd.

import zipfile
from django.core.files import File

f='s1.zip'
zf = zipfile.ZipFile(f, "w")
for dirname, subdirs, files in os.walk(src):
    zf.write(dirname)
    for filename in files:
        zf.write(os.path.join(dirname, filename))
        zf.open(os.path.join(dirname, filename))
newobj.fd= File(zf)

I do this thing for a text file and it works:

f=file('text.txt')
newobj.fd2=File(f)
f.close()

How do the same thing for a zipfile?

Answer 1

import zipfile
from django.core.files import File

f='s1.zip'
zf = zipfile.ZipFile(f, "w")
for dirname, subdirs, files in os.walk(src):
    zf.write(dirname)
    for filename in files:
        zf.write(os.path.join(dirname, filename))
zf.close()
newobj.fd = File(f)

What does this give?