How to handle mainloop and serve_forever socket server together in if __name__ == '__main_

I am using Tkinter which has a QRcode generate button. I want to create a QRcode based on the provided URL and if I click the QRcode generate button then it will generate a QRcode and the URL will be active forever. The code I tried so far.

generate_button = tk.Button(my_w,font=22,text='Generate QR code', command=lambda:my_generate())
generate_button.place(relx=0.2, rely=0.5, anchor=CENTER)

qrcode_label=tk.Label(my_w)
qrcode_label.place(relx=0.6, rely=0.5, anchor=CENTER)

link ='http://192.x.x.x:8010'
PORT = 8010

def my_generate():
    global my_img
    my_qr = pyqrcode.create(link) 
    my_qr = my_qr.xbm(scale=10)
    my_img=tk.BitmapImage(data=my_qr)
    qrcode_label.config(image=my_img)

So far everything is cool. Now if I try to activate the server beside the main Tkinter window, seems both two loops are going to conflict and the application gets crashed.

if __name__ == '__main__':
    Handler = http.server.SimpleHTTPRequestHandler
    httpd = socketserver.TCPServer(("", PORT), Handler)
    print("serving at port", PORT)
    httpd.serve_forever()
    my_w.mainloop()

Tried some ways but nothing helps me so far.

CodePudding user response：

To avoid the conflict between the Tkinter mainloop and the server's loop, you can use the threading module to run the server on a separate thread. You can create a new thread and run the server's serve_forever() method on that thread. Here is an example of how you can do this:

import threading

if __name__ == '__main__':
    my_w.mainloop()
    server_thread = threading.Thread(target=serve_forever)
    server_thread.start()

def serve_forever():
    Handler = http.server.SimpleHTTPRequestHandler
    httpd = socketserver.TCPServer(("", PORT), Handler)
    print("serving at port", PORT)
    httpd.serve_forever()

This way, the Tkinter mainloop and the server's loop will run in parallel and won't conflict with each other.