如何在 Python 的單獨線程中啟動 flask api 服務器

Question

我正在開發一個 python 項目，其中包括基於pyqt5和flask的 ZDB974238714CA8DE1434A7CE1D08A3。 在 UI 中，有start和stop按鈕，按下該按鈕將啟動flask api 服務器並在單擊停止按鈕時停止。

from server import start_local_server
from multiprocessing import Process

"""
SOME CODE
"""

def start_server_btn_event(self):
    p1 = Process(target=start_local_server())
    p1.start()
    

def stop_server_btn_event(self):
    # Code to stop the api server

但上述操作會使整個 UI 無響應，我無法單擊 UI 上的任何其他 object。 如何在單獨的線程或進程中單擊按鈕時運行 api 服務器，以便其他 UI 對象處於活動狀態並可以執行其功能。 謝謝

最小可重現示例：

app.py：這包含啟動本地服務器的 pyqt5 按鈕

import sys
from PyQt5.QtWidgets import QApplication, QWidget, QPushButton
from server import start_local_server
from PyQt5.QtCore import pyqtSlot
from multiprocessing import Process


class App(QWidget):

    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):
        # Start server here
        p1 = Process(target=start_local_server())
        p1.start()


    @pyqtSlot()
    def on_click_stop_btn(self):
        print("Stop server ")

    @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_())

server.py：這是 flask api 服務器代碼

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)

Answer 1

我已經通過啟動一個單獨的線程來運行 API 服務器解決了這個問題：

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()