How to get the filepath of a ZipFile object?

Question

If I have a zipfile.ZipFile object, how can I determine the file path of the zip file from the object?

a = zipfile.ZipFile('C:\\path\\zipfile.zip')

a.get_file_path()

where get_file_path() should return 'C:\\path\\zipfile.zip'.

Answer 1

You can access it using filename attribute:

a.filename