2

我有以下 python (pygtk) 程序。当我将鼠标移到显示的菜单上时，如果单击托盘中的项目，复选框将被选中并立即再次取消选中。我使用的是 Ubuntu 10.10 或 11.04。

#!/usr/bin/python

import  gtk
import  glib
import  subprocess
import  time
import  sys



class StatusIcon:
    def __init__(self):
        self.statusicon = gtk.StatusIcon()
        self.statusicon.set_from_stock(gtk.STOCK_HOME) 
        self.statusicon.connect("popup-menu", self.right_click_event)



    def right_click_event(self, icon, button, time):
        """
        We show a menu
        """
        menu = gtk.Menu()

        submenu = gtk.Menu()
        menuitem    = gtk.MenuItem("1")
        submenu.append(menuitem)
        menuitem    = gtk.MenuItem("2")
        submenu.append(menuitem)

        lst = ["a","b","c"]

        for item in lst:
            newmenuitem = gtk.CheckMenuItem(str(item))
            newmenuitem.set_submenu(submenu)
            menu.append(newmenuitem)

        # Now add all other menu stuff
        menu.append(gtk.SeparatorMenuItem())
        menuexit    = gtk.CheckMenuItem("exit")
        menuexit.connect("button-press-event", self.exit)
        menu.append(menuexit)

        # Show the menu
        menu.show_all()
        menu.popup(None, None, gtk.status_icon_position_menu, button, time, self.statusicon)

    def exit(self, widget, event):
        """
        Menu exit pressed
        """
        if event.button == 1:   #LEFT
            print   "terminate"
            gtk.main_quit()


StatusIcon()
gtk.main()
4

1 回答 1

2

子菜单在出现时“切换”复选框的状态。您可以通过处理复选框菜单项的切换信号来对抗这种行为。

class StatusIcon:

...

  def callback(self, widget, data=None):
    widget.set_active(False)

...

    for item in lst:
        newmenuitem = gtk.CheckMenuItem(str(item))
        newmenuitem.connect("toggled", self.callback, "")
于 2011-06-17T20:16:12.923 回答