Python：如何为GUI类创建单独的模块

此代码运行良好。MyApp是完成所有工作的类，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，以便创建一个模块文件GUI.py并将其导入MyApp文件：

from gui import *

---

就这样。我挣扎的地方：gui.py看起来怎么样？MyGUI（）如何访问MyApp方法？这样做明智吗？关于构造的任何其他建议？

gui.py文件只需

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) 

将导入添加到myapp.py的顶部，一切都会正常工作

---

如果有意义的话，我会尝试将代码分成不同的文件。它使阅读更加清晰。

在您的

gui.py

from myapp import MyApp

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

    app = MyApp()

    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)  

并在您的

myapp.py

（注意第一行），确保两个文件位于同一目录中，否则必须相对更改导入

from gui import MyGUI

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")

---

最后，我做到了这一点——这似乎是在应用程序和gui之间实现清晰分离和通信的最佳方法

图形用户界面：

应用程序：

---

是的，把他们俩分开没关系。而且，通过import all“gui.py”第11行的

导入后端方法很容易。name错误：名称“app”未定义我已经更新了代码，但是您的示例似乎在两个类之间有很多交错。但是，您必须始终尝试实现类之间的单向依赖。因此，当您编写代码时，请尝试在一个类中编写所有后端方法，并将整个类导入到前端文件中。不要将前端方法带入后端类。谢谢，但仍然不起作用。>ImportError:无法导入名称“MyGUI”，因此至少无法导入pyhton 3.6。禁止相互参照。您的评论很有道理，但如果您想严格地将GUI与核心分离，这就是本质。GUI向core发送消息并从core接收消息。非常开放的通用方法来解决这个问题。
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])

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 ...