python - How to stream from ZipFile? How to zip "on the fly"?

Question

I want to zip a stream and stream out the result. I'm doing it using AWS Lambda which matters in sense of available disk space and other restrictions. I'm going to use the zipped stream to write an AWS S3 object using upload_fileobj() or put(), if it matters.

I can create an archive as a file until I have small objects:

import zipfile
zf = zipfile.ZipFile("/tmp/byte.zip", "w")
zf.writestr(filename, my_stream.read())
zf.close()

For large amount of data I can create an object instead of file:

from io import BytesIO
...
byte = BytesIO()
zf = zipfile.ZipFile(byte, "w")
....

but how can I pass the zipped stream to the output? If I use zf.close() - the stream will be closed, if I don't use it - the archive will be incomplete.

Answer 1

You might like to try the zipstream version of zipfile. For example, to compress stdin to stdout as a zip file holding the data as a file named TheLogFile using iterators:

#!/usr/bin/python3
import sys, zipstream
with zipstream.ZipFile(mode='w', compression=zipstream.ZIP_DEFLATED) as z:
    z.write_iter('TheLogFile', sys.stdin.buffer)
    for chunk in z:
        sys.stdout.buffer.write(chunk)