Using UploadFile for direct file upload without a multipart/form-data request

Question

I have an FastAPI endpoint for handling file uploads that looks something like this:

@app.post('/upload')
async def accept_some_file(f: UploadFile):
    content = await f.read()
    # ... do stuff with content and generate a response

but this appears to only work with multipart/form-data encoded payloads.

I'd like to be able to send file bytes directly through a request that looks like this:

POST /upload HTTP/1.1
Host: localhost:8080
User-Agent: curl/7.79.1
Accept: */*
Content-Type: image/jpeg
Content-Length: 11044

... image bytes

Is there a FastAPI setting I can use to allow this? Or is there another request type that makes more sense for this use case?

This question helps you? https://stackoverflow.com/questions/63048825/how-to-upload-file-using-fastapi — Gonzalo Odiard, Apr 29 '22 at 18:37
No, that question involves using `multipart/form-data` as the upload encoding. I would like to avoid that. — kingkupps, Apr 29 '22 at 18:39

score 1 · Accepted Answer · answered Apr 29 '22 at 19:21

You can get the request body as bytes using await request.body():

from fastapi import Request

@app.post('/upload')
async def accept_some_file(request: Request):
    body = await request.body()

or, you can also access the request body as a stream (in this way the byte chunks are provided without storing the entire body to memory) - see Starlette documentation. Example:

@app.post('/upload')
async def accept_some_file(request: Request):
    body = b''
    async for chunk in request.stream():
        body += chunk

Using UploadFile for direct file upload without a multipart/form-data request

1 Answers1