Sending Image in Multipart to FastAPI Endpoint from Python

The code is in the gist below:

It is quite straightforward, isn’t it? When you are preparing input instead of file_path you can also give name of the file or any default value as well. In that case, the input changes as:

files = {‘file’: (“some_useful_name”, open(file_path, ‘rb’), “image/jpeg”)}

That’s it!