Concurrently start function and return early [duplicate]

Question

I need to return a response from my FastAPI path operation, but before this I want to send a slow request and I don't need to wait for result of that request, just log errors if there are any. Can I do this by means of Python and FastAPI? I would not like to add Celery to the project.

Here is what I have so far, but it runs synchronously:

import asyncio
import requests


async def slow_request(data):
    url = 'https://external.service'
    response = requests.post(
        url=url,
        json=data,
        headers={'Auth-Header': settings.API_TOKEN}
    )
    if not response.status_code == 200:
        logger.error('response:', response.status_code)
        logger.error('data', data)


@router.post('/order/')
async def handle_order(order: Order):
    json_data = {
        'order': order
    }
    
    task = asyncio.create_task(
        slow_request(json_data)
    )
    await task

    return {'body': {'message': 'success'}}

Answer 1

OK, if nobody wants to post an answer here are the solutions:

Solution #1

We can just remove await task line as alex_noname suggested. It will work because create_task schedules task and we are no longer awaiting for its completion.

@router.post('/order/')
async def handle_order(order: Order):
    json_data = {
        'order': order
    }
    
    task = asyncio.create_task(
        slow_request(json_data)
    )

    return {'body': {'message': 'success'}}

Solution #2

I ended up with BackgroundTasks as HTF suggested as I'm already using FastAPI anyway, and this solution seems more neat to me.

@router.post('/order/')
async def handle_order(order: Order, background_tasks: BackgroundTasks):
    json_data = {
        'order': order
    }
    
    background_tasks.add_task(slow_request, json_data)

    return {'body': {'message': 'success'}}

This even works without async before def slow_request(data):