How to read zipfile from stdin

I'm trying to solve reading a zipfile from stdin in python, but I keep getting issues. What I want is to be able to run cat test.xlsx | python3 test.py and create a valid zipfile.ZipFile object without first writing a temporary file if possible.

My initial approach was this, but ZipFile complained the file is not seekable,

import sys
import zipfile

zipfile.ZipFile(sys.stdin)

so I changed it around, but now it complains that this is not a valid zip file:

import io
import sys
import zipfile

zipfile.ZipFile(io.StringIO(sys.stdin.read()))

Can this be solved without writing the zip to a temporary file?

Zip files are binary data, not UTF-8 encoded text. You won't be able to read the file into a str with sys.stdin.read() without immediately hitting a UnicodeDecodeError: 'utf-8' codec can't decode byte ... error.

Instead, you can access the underlying binary buffer object to read stdin as raw bytes. Pair that with BytesIO to get an in-memory seekable file-like object:

zipfile.ZipFile(io.BytesIO(sys.stdin.buffer.read()))

Alternatively, if you provide a seekable stdin (for example, by redirecting stdin instead of streaming from a pipe), you can operate on sys.stdin.buffer directly:

zipfile.ZipFile(sys.stdin.buffer)

paired with something like

python3 test.py <test.xlsx

If you care to, you can select between the two depending on whether stdin is seekable by querying the IO object's seekable method:

if sys.stdin.buffer.seekable():
    zip_file = zipfile.ZipFile(sys.stdin.buffer)
else:
    buffer = io.BytesIO(sys.stdin.buffer.read())
    zip_file = zipfile.ZipFile(buffer)

print(zip_file.filelist)