两个线程：1 - Flask 应用程序，2 - 工作网络摄像头

Question

我有一个线程用于不断处理来自相机的图像，第二个线程是闪存服务器。来自第一个线程的帧必须传输到服务器。但两个线程都不会为我启动 - 无论是第一个还是第二个。

#!/usr/bin/env python
from datetime import datetime
from flask import Flask, render_template, Response, request
import numpy as np
import cv2
from threading import Thread

# Initialize the Flask app
app = Flask(__name__)
# global pwm_az pwm_tilt
pwm_az = 0
pwm_tilt = 0
global frame
frame = 0

def vd():
    global frame
    # define a video capture object
    vid = cv2.VideoCapture(0)
    print("Запущен поток с камерой")
    th = Thread(target=webWork(),args=())
    th.start()
    th.join()

    while (True):
        
        # Capture the video frame
        # by frame
        ret, frame = vid.read()

        # Display the resulting frame
        cv2.imshow('frame', frame)

        # the 'q' button is set as the
        # quitting button you may use any
        # desired button of your choice
        if cv2.waitKey(1) & 0xFF == ord('q'):
            break

    # After the loop release the cap object
    vid.release()
    # Destroy all the windows
    cv2.destroyAllWindows()
def gen_frames():
    global frame
    yield (b'--frame\r\n'
                   b'Content-Type: image/jpeg\r\n\r\n' + frame + b'\r\n')


@app.route('/')
def index():
    return render_template('index_2.html')

@app.route('/video_feed')
def video_feed():
    return Response(gen_frames(), mimetype='multipart/x-mixed-replace; boundary=frame')

def webWork():
    if __name__ == "__main__":
        app.run(host="localhost", port=8086, debug=True)


thread_videoWork = Thread(target=vd(), args=())
thread_webWork = Thread(target=webWork(), args=())
thread_webWork.start()
thread_videoWork.start()

对我来说很重要的是，从相机进行处理的过程是在脚本开始时进行的，而不是在将其成像到服务器时开始的。

尝试创建第三个线程没有帮助。

Answer 1

我如何找到它来修复它。有必要将这两个任务移至单独的线程中：

if __name__ == "__main__":
    try:
        print(f'start first thread')
        t1 = Thread(target=runApp).start()
        print(f'start second thread')
        t2 = Thread(target=videoWork).start()
    except Exception as e:
        print("Unexpected error:" + str(e))