如何通过 python 请求库将 Starlette FormData 数据结构发送到 FastAPI 端点

Question

My system architecture currently sends a form data blob from the frontend to the backend, both hosted on localhost on different ports. The form data is recieved in the backend via the FastAPI library as shown.

@app.post('/avatar/request')
async def get_avatar_request(request: Request, Authorize: AuthJWT = Depends()):
    form = await request.form()
    return run_function_in_jwt_wrapper(get_avatar_output, form, Authorize, False)

Currently, I am trying to relay the form data unmodified to another FASTApi end point from the backend using the request library, as follows:

response = requests.post(models_config["avatar_api"], data = form_data, headers = {"response-type": "blob"})

While the destination endpoint does receive the Form Data, it seemed to not have parsed the UploadFile component properly. Instead of getting the corresponding starlette UploadFile data structure, I instead receive the string of the classname, as shown in this error message:

FormData([('avatarFile', '<starlette.datastructures.UploadFile object at 0x7f8d25468550>'), ('avatarFileType', 'jpeg'), ('background', 'From Image'), ('voice', 'en-US-Wavenet-B'), ('transcriptKind', 'text'), ('translateToLanguage', 'No translation'), ('transcriptText', 'do')])

How should I handle this problem?

Answer 1

FileUpload is a python object, you'd need to serialize it somehow before using requests.post() then deserialize it before actually getting the content out of it via content = await form["upload-file"].read() . I don't think you'd want to serialize a FileUpload object though (if it is possible), rather you'd read the content of the form data and then post that.

Even better, if your other FastAPI endpoint is part of the same service, you might consider just calling a function instead and avoid requests altogether (maybe use a controller function that the route function calls in case you also need this endpoint to be callable from outside the service, then just call the controller function directly avoiding the route and the need for requests). This way you can pass whatever you want without needing to serialize it.

If you must use requests, then I'd read the content of the form then create a new post with with that form data. eg

form = await request.form()  # starlette.datastructures.FormData
upload_file = form["upload_file"]  # starlette.datastructures.UploadFile - not used here, but just for illustrative purposes
filename = form["upload_file"].filename  # str
contents = await form["upload_file"].read()  # bytes
content_type = form["upload_file"].content_type  # str
...
data = {k: v for k, v in form.items() if k != "upload-file"} # the form data except the file
files = {"upload-file": (filename, contents, content_type)} # the file
requests.post(models_config["avatar_api"], files=files, data=data, headers = {"response-type": "blob"})