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Example Code

from fastapi.applications import FastAPI
import uvicorn
from fastapi.requests import Request
from fastapi.responses import StreamingResponse
from visitor_camera_fastapi import VideoCameraV


app = FastAPI()


async def genv(camera, request):
    while True:
        print(f'is disconnect {request._is_disconnected}')
        frame = camera.get_frame()
        yield (b'--frame\r\n'
            b'Content-Type: image/jpeg\r\n\r\n' + frame + b'\r\n\r\n')

@app.get("/getCameraStream")
async def detectEmployees(request:Request):
    # body = await request.body()
    query = request.query_params
    return StreamingResponse(genv(VideoCameraV(request)), media_type='multipart/x-mixed-replace; boundary=frame')

Description

hi, In my project we capture the stream in the backend and passing to the front-end through API. If the front-end stop getting the stream .then also in the backend stream is getting pass. how Can I stop this stream

Operating System

Windows

Operating System Details

No response

FastAPI Version

0.63.0

Python Version

3.7

Additional Context

I am getting the socket.send() error .there is way I can check the client is disconnect . I have tried with request._is_disconnected also but it returing false if the client disconnected also