creating a python zip without any directory info

Question

I'm attempting to create a zip file to upload to AWS Lambda, but I can't seem to create the zip file correctly.

What I need to create is a zip file that only contains the files themselves, using this code:

def zipdir(path, ziph):
    for root, dirs, files in os.walk(path):
        for file in files:
            filepath = os.path.join(root, file)
            if file == 'config.json' or file.startswith('.'):
                pass
            else:
                info = zipfile.ZipInfo(filename=file)
                ziph.write(filename=filepath, arcname='./' + file)

I successfully create the archive, but I always get the file contents in a directory with the same name as the archive.

If I compress:

code.js

And then decompress, testing.zip I'll get:

testing/code.js

How do I prevent the directory from being created?

Answer 1

Not familiar with python, but I use Makefile instead

create:
    [ -e "pkg.zip" ] && rm pkg.zip &2> /dev/null
    zip -r pkg.zip . \
        --exclude=test* \
        --exclude=node_modules/aws-sdk*
    aws lambda create-function \
        --function-name $(func) \
        --zip-file fileb://pkg.zip \
        --role arn:aws:iam::YOUR_ACCOUNT_ID:role/gff_lambda \
        --handler index.handler \
        --timeout 30 \
        --runtime nodejs