Python Tkinter 菜单命令不起作用

2022-01-23 00:00:00 python tkinter dictionary command

问题描述

我正在尝试在 Python 2.6.5 中执行以下代码.我想要做的是显示一个带有应用程序"菜单的主窗口.我希望菜单有一系列命令，这些命令应该对应于 Apps 字典的键.当我单击该命令时，我希望默认 Web 浏览器打开并导航到该特定键的 Apps 字典中的 url.相反，当我执行代码时，浏览器将打开到 Apps 字典中的第一个 url，而无需任何点击.请帮忙！

I am trying to execute the following code in Python 2.6.5. What I want to do is show a main window with an 'Applications' menu. I want the menu to have a series of commands which should correspond to the keys of the Apps dictionary. When I click the command, I would like the default web browser to open and navigate to the url in the Apps dictionary for that particular key. Instead, when I execute the code the browser is opening to the first url in the Apps dictionary without any clicking. Help please!

from Tkinter import *
import webbrowser

#Real links are to pages on Intranet.
Apps={
     'Google':'http://www.google.com/',
     'Yahoo':'http://www.yahoo.com/'
     }

def openApp(appURL):
     webbrowser.open(appURL, new=1, autoraise=1)
     return None

root=Tk()
menubar=Menu(root)
root.config(menu=menubar)
appsMenu=Menu(menubar)
for app in Apps:
     appsMenu.add_command(label=app, command=openApp(Apps[app]))
menubar.add_cascade(label='Apps', menu=appsMenu)
root.mainloop()

解决方案

 appsMenu.add_command(label=app, command=openApp(Apps[app]))

调用函数的命令参数需要包装在 lambda 中，以防止它们被立即调用.此外，在 for 循环中绑定的命令需要循环变量作为默认参数，以便每次都绑定正确的值.

Command parameters that call functions need to be wrapped in a lambda, to prevent them from being called right away. Additionally, commands bound within a for loop need the looping variable as a default argument, in order for it to bind the right value each time.

 appsMenu.add_command(label=app, command=lambda app=app: openApp(Apps[app]))

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