openai library in Python not able to parse URLs to search using keywords

Question

I have written a short chatbot prototype, using the OpenAI library, that takes a text input from the keyboard, runs that search through a list of websites, processes any returned text via ChatGPT and gives it back to the user.

I am using the following code to create a response

html = openai.File.create(url=source).text

Where source is created by iterating through a list of URLs.

I am building using a standard Raspberry Pi, Thonny is my IDE, Python 3.9.2.

I get the following errors. It seems to me that the openai.File.create method does not accept 'url', but I cannot find a list pf the acceptable keyword arguments in the Library documentation.

Do you how to use a keyword which will take a URL from my list per the above?

Many thanks

Error searching source https://en.wikipedia.org/wiki/Python_(programming_language): create() got an unexpected keyword argument 'url' Error searching source https://docs.python.org/3/tutorial/index.html: create() got an unexpected keyword argument 'url' Error searching source https://realpython.com/tutorials/: create() got an unexpected keyword argument 'url' Error searching source https://www.geeksforgeeks.org/python-programming-language/: create() got an unexpected keyword argument 'url' Error searching source https://www.python.org/: create() got an unexpected keyword argument 'url' Error searching source https://positivepython.co.uk: create() got an unexpected keyword argument 'url'

Answer 1

All the examples I've seen of File.create take in a file, not an url. The keyword is file. E.g.

openai.File.create(file=open("myfile.jsonl"), purpose="search")

according some docs the File.create method creates files.

Here is the source code for the File class create method:

@classmethod
def create(
    cls,
    file,
    purpose,
    model=None,
    api_key=None,
    api_base=None,
    api_type=None,
    api_version=None,
    organization=None,
    user_provided_filename=None,
):
    requestor, url, files = cls.__prepare_file_create(
        file,
        purpose,
        model,
        api_key,
        api_base,
        api_type,
        api_version,
        organization,
        user_provided_filename,
    )
    response, _, api_key = requestor.request("post", url, files=files)
    return util.convert_to_openai_object(
        response, api_key, api_version, organization
    )

as you can see, no url keyword.