适用于Python3.6的Zipfile模块：写字节而不是Odoo文件

Question

I've been trying to use the zipfile module for Python 3.6 to create a .zip file which contains multiple objects.
My problem is, I have to manage files from an Odoo database that only allows me to use bytes objects instead of files.

This is my current code:

import zipfile

empty_zip_data = b'PK\x05\x06\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00'
zip = zipfile.ZipFile(empty_zip_data, 'w')

# files is a list of tuples: [(u'file_name', b'file_data'), ...]
for file in files:
    file_name = file[0]
    file_data = file[1]
    zip.writestr(file_name, file_data)

Which returns this error:

File "/Library/Frameworks/Python.framework/Versions/3.6/lib/python3.6/zipfile.py", line 1658, in writestr
  with self.open(zinfo, mode='w') as dest:
File "/Library/Frameworks/Python.framework/Versions/3.6/lib/python3.6/zipfile.py", line 1355, in open
  return self._open_to_write(zinfo, force_zip64=force_zip64)
File "/Library/Frameworks/Python.framework/Versions/3.6/lib/python3.6/zipfile.py", line 1468, in _open_to_write
  self.fp.write(zinfo.FileHeader(zip64))
File "/Library/Frameworks/Python.framework/Versions/3.6/lib/python3.6/zipfile.py", line 723, in write
  n = self.fp.write(data)
AttributeError: 'bytes' object has no attribute 'write'

How am I supposed to do it? I followed the ZipFile.writestr() docs , but that got me nowhere...

EDIT: using file_data = file[1].decode('utf-8') as second parameter is not useful either, I get the same error.

Answer 1

As mentioned in my comment, the issue is with this line:

empty_zip_data = b'PK\x05\x06\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00'
zip = zipfile.ZipFile(empty_zip_data, 'w')

You're trying to pass a byte object into the ZipFile() method, but like open() it is expecting a path-like object.

In your case, you might want to utilize the tempfile module (in this particular example we'll use SpooledTemporaryFile from this relevant question :

import tempfile
import zipfile

# Create a virtual temp file
with tempfile.SpooledTemporaryFile() as tp:

    # pass the temp file for zip File to open
    with zipfile.ZipFile(tp, 'w') as zip:
        files = [(u'file_name', b'file_data'), (u'file_name2', b'file_data2'),]
        for file in files:
            file_name = file[0]
            file_data = file[1]
            zip.writestr(file_name, file_data)

    # Reset the cursor back to beginning of the temp file
    tp.seek(0)
    zipped_bytes = tp.read()

zipped_bytes
# b'PK\x03\x04\x14\x00\x00\x00\x00\x00\xa8U ... \x00\x00'

Note the use of context managers to ensure all your file objects are closed properly after being loaded.

This gives you zipped_bytes which is the bytes you want to pass back to Odoo. You can also test the zipped_bytes by writing it to a physical file to see what it looks like first:

with open('test.zip', 'wb') as zf:
    zf.write(zipped_bytes)

If you are handling file size that are considerably large, make sure to pay attention and make use of max_size argument in the documentation.

Answer 2

If you want to handle all of this in memory without a temporary file then use io.BytesIO as the file object for ZipFile :

import io
from zipfile import ZIP_DEFLATED, ZipFile

file = io.BytesIO()
with ZipFile(file, 'w', ZIP_DEFLATED) as zip_file:
    for name, content in [
        ('file.dat', b'data'), ('another_file.dat', b'more data')
    ]:
        zip_file.writestr(name, content)

zip_data = file.getvalue()
print(zip_data)

You may also want to set the compression algorithm as shown because otherwise the default (no compression!) is used.