Right Click Menu (context menu) using PyGTK

Question

So I'm still fairly new to Python, and have been learning for a couple months, but one thing I'm trying to figure out is say you have a basic window...

#!/usr/bin/env python

import sys, os
import pygtk, gtk, gobject

class app:

   def __init__(self):
    window = gtk.Window(gtk.WINDOW_TOPLEVEL)
    window.set_title("TestApp")
    window.set_default_size(320, 240)
    window.connect("destroy", gtk.main_quit)
    window.show_all()

app()
gtk.main()

I wanna right click inside this window, and have a menu pop up like alert, copy, exit, whatever I feel like putting down.

How would I accomplish that?

First off, you should read up on how GTK utilizes events. Start with getting right-clicking to do something, then read up on how to do context menus. I've never used PyGTK or GTK, but there's probably a widget for it. — JAB
– JAB, Commented Jul 7, 2011 at 20:43

James Hurford · Accepted Answer · 2011-08-23 22:57:59Z

13

There is a example for doing this very thing found at http://www.pygtk.org/pygtk2tutorial/sec-ManualMenuExample.html

It shows you how to create a menu attach it to a menu bar and also listen for a mouse button click event and popup the very same menu that was created.

I think this is what you are after.

EDIT: (added further explanation to show how to respond to only right mouse button events)

To summarise.

Create a widget to listen for mouse events on. In this case it's a button.

button = gtk.Button("A Button")

Create a menu

menu = gtk.Menu()

Fill it with menu items

menu_item = gtk.MenuItem("A menu item")
menu.append(menu_item)
menu_item.show()

Make the widget listen for mouse press events, attaching the menu to it.

button.connect_object("event", self.button_press, menu)

Then define the method which handles these events. As is stated in the example in the link, the widget passed to this method is the menu that you want popping up not the widget that is listening for these events.

def button_press(self, widget, event):
    if event.type == gtk.gdk.BUTTON_PRESS and event.button == 3:
        #make widget popup
        widget.popup(None, None, None, event.button, event.time)
        pass

You will see that the if statement checks to see if the button was pressed, if that is true it will then check to see which of the buttons was pressed. The event.button is a integer value, representing which mouse button was pressed. So 1 is the left button, 2 is the middle and 3 is the right mouse button. By checking to see if the event.button is 3, you are only responding to mouse press events for the right mouse button.

edited Aug 23, 2011 at 22:57

answered Jul 7, 2011 at 22:07

James Hurford

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7 Comments

Michael Schwartz Over a year ago

Thanks! That's almost exactly what I've been trying to figure out. Only problem though that the menu shows on left click, as well as right. In which I don't want it to show up on left click, but I should be able to fix that. Thanks!

neydroydrec Over a year ago

@mikethedj4: could you please tell if/how you achieved to define the click triggering the pop-up menu? Thanks.

James Hurford Over a year ago

@Benjamin: I think the answer should now cover what you are after

Michael Schwartz Over a year ago

Yes, and sorry for the long reply.

jackw11111 Over a year ago

popup prototype now looks like popup(parent_menu_shell, parent_menu_item, func, data, button, activate_time).

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