如何在 flask API Python 中移出 app.run()

Question

I am working on python project which includes starting a flask API server on a button click. For buttons and UI I am using pyqt5 . So on start server button click flask api server will start and on stop server button click, flask api server will stop.

I am able to start the server but the problem is when we start the flask api server, we use app.run(HOST, 80) . During this the control always remains here and doesnt move out of it due to which when I click stop button, it doesnt stop. Below is the code:

app.py

import sys
import time
from PyQt5.QtWidgets import QApplication, QWidget, QPushButton
from server import start_local_server
from PyQt5.QtCore import pyqtSlot
from threading import Thread


run = True


def start_api_server():
    while run:
        start_local_server()
        print("server is running")
    time.sleep(1)
    print("SERVER HAS STOPPED")


class App(QWidget):
    global run

    def __init__(self):
        super().__init__()
        self.title = 'PyQt5 button - pythonspot.com'
        self.left = 10
        self.top = 10
        self.width = 320
        self.height = 200
        self.initUI()

    def initUI(self):
        self.setWindowTitle(self.title)
        self.setGeometry(self.left, self.top, self.width, self.height)

        start_btn = QPushButton('Start Server', self)
        start_btn.move(100, 70)
        start_btn.clicked.connect(self.on_click_start_btn)

        stop_btn = QPushButton('Stop Server', self)
        stop_btn.move(200, 70)
        stop_btn.clicked.connect(self.on_click_stop_btn)

        fun_btn = QPushButton('Click to check responsiveness', self)
        fun_btn.move(150, 100)
        fun_btn.clicked.connect(self.on_click_fun_btn)

        self.show()

    @pyqtSlot()
    def on_click_start_btn(self):
        Thread(target=start_api_server).start()
        
    @pyqtSlot()
    def on_click_stop_btn(self):
        print("Stop server ")
        run = False
        
    @pyqtSlot()
    def on_click_fun_btn(self):
        print('If it is working, this means UI is responsive')


if __name__ == '__main__':
    app = QApplication(sys.argv)
    ex = App()
    sys.exit(app.exec_())

In above code, I have button click function on_click_start_btn this starts a thread with target start_api_server . This will further start the local flask server. When I want to stop the server, I will simply make run as False which will break the start_api_server function and the server will stop. As per my understanding this should work fine.

Below is the code for flask api:

server.py

import os
import datetime
from flask import Flask, jsonify
from flask_cors import CORS


app = Flask(__name__)
CORS(app)
wsgi_app = app.wsgi_app


@app.route('/api/status')
def check_status():
    return jsonify({'status': 'ok', 'date': datetime.datetime.now().isoformat()}), 200


def start_local_server():
    HOST = os.environ.get('SERVER_HOST', 'localhost')
    try:
        PORT = int(os.environ.get('SERVER_PORT', '5555'))
    except ValueError:
        PORT = 5555
    app.run(HOST, 80)

The problem occurs in above flask code. In above code, there is function start_local_server() which we are using in app.py . In this we have app.run() , when our code reaches here, it always remains here and never moves out it due to which I am not able to stop this server.

I want to simply make code where I can start and stop the flask server using button click but due to app.run , its not working. Can anyone please help me with this issue. Is there any alternative to above problem. Please help. Thanks

Updated code as per answer:

server = Process

def start_local_server():
    global server
    server = Process(target=app.run, args=('localhost', 80))
    # HOST = os.environ.get('SERVER_HOST', 'localhost')
    # try:
    #     PORT = int(os.environ.get('SERVER_PORT', '5555'))
    # except ValueError:
    #     PORT = 5555
    # app.run(HOST, 80)


def stop_local_server():
    global server
    server.terminate()

Answer 1

You can run app.run() in a seperate Process . Multiprocessing would be helpful in your case, since you're using pyQt.

from multiprocessing import Process

server = Process(target=app.run, args=(HOST, 80))
server.start() # to start the server
server.terminate() # to terminate the server

You can take a deeper look at multiprocessing library documentation here https://docs.python.org/2/library/multiprocessing.html

You forgot to strart the server:)

server = None

def start_local_server():
    global server
    server = Process(target=app.run, args=('localhost', 80))
    server.start()
    # HOST = os.environ.get('SERVER_HOST', 'localhost')
    # try:
    #     PORT = int(os.environ.get('SERVER_PORT', '5555'))
    # except ValueError:
    #     PORT = 5555
    # app.run(HOST, 80)


def stop_local_server():
    global server
    server.terminate()

Answer 2

I have resolved the issue by starting the API server in separate thread so that Ui keeps on working. Then I am shutting down the API server using shutdown api

run = True


def start_api_server():
    while run:
        start_local_server()
        
    time.sleep(1)
    print("SERVER HAS STOPPED")

@pyqtSlot()
def on_click_start_btn(self):
    Thread(target=start_api_server).start()

Above is the code to start the server on button click. Below is how I am shutting it down:

@pyqtSlot()
def on_click_stop_btn(self):
    global run
    print("STOPPING SERVER")
    run = False
    try:
        x = requests.get("http://localhost:80/api/shutdown")
        rdata = x.text
        rdata = json.loads(rdata)
        if rdata['status']:
            print("Server shutting down")
        print(rdata)

    except Exception as e:
        print(e)

Above code calls a shutdown API which terminate the flask api server. Below is the code for it:

def shutdown_server():
    func = request.environ.get('werkzeug.server.shutdown')
    if func is None:
        raise RuntimeError('Not running with the Werkzeug Server')
    func()


@app.route('/api/shutdown')
def shutdown():
    shutdown_server()
    return 'Server shutting down...'