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()
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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