Question

此代码运行正常。 MyApp是完成所有工作的类，而MyGUI是显示和请求MyApp数据的用户界面。

class MyGUI(): # displays results from MyApp and sends request to MyApp (e.g. fetch prices new prices)

    def __init__(self):
        print("GUI running")

    def user_request_price(self,ticker):        
        self.req_price(ticker)

    # methods I request from MyApp 
    def req_price(self,ticker): 
        app.get_price(ticker)

    # methods I receive from MyApp
    def print_price(self,val,price):
        print (val,":",price)    

class MyApp(): # does a lot of stuff, e.g. fetch prices from a server

    def __init__(self):
        self.id = 0
        self.gui = MyGUI() # start gui

    # methods called by GUI
    def get_price(self, ticker):
        if ticker == "MSFT": price = 20.23
        self.output_price(ticker,price)

    # methods sent to GUI
    def output_price(self,ticker,price):
        self.gui.print_price(ticker,price)


if __name__ == "__main__": 
    app = MyApp()
    app.gui.user_request_price("MSFT")

现在，我想将GUI放入一个单独的模块中，因此创建一个模块文件gui.py并将其导入MyApp文件中：

from gui import *

就是这样。我在哪里挣扎：gui.py的外观如何以及MyGUI（）如何访问MyApp方法？进行这种分离是否明智？还有其他结构建议吗？

Answer 1

最后我做到了-似乎是在app和gui之间进行清晰分离和通信的最佳方法。

Gui：

import queue

def __init__(self):
    threading.Thread.__init__(self)
    self.requests = queue.Queue()  # request queue for App
    self.start()

def queue_request(self,reqId,val):
    self.requests.put([reqId,val])

APP：

import threading
import queue

def checkGUIQueue(self):
    threading.Timer(1.0, self.checkGUIQueue).start() # check every 1 second                
    while not self.gui.requests.empty():
        (id,value) = self.gui.requests.get()
        ... process request ...

Python：如何为GUI类创建单独的模块

1 个答案: