[gnome-shell] App menu: only create the popup menu with a GMenu

[Giovanni Campagna <gcampagna src gnome org>]

commit 1bac40fbe34b29b021075be4519aafdcfa6f2a13
Author: Giovanni Campagna <gcampagna src gnome org>
Date:   Mon Jan 9 16:31:44 2012 +0100

    App menu: only create the popup menu with a GMenu
    
    Since the application proxy is created asynchronously, at the time
    the GActionGroup (GActionMuxer) is created, there is no GDBusMenu yet.
    Defer creating the menu in that case.
    Also, clear out signal handlers if have no target application.
    
    https://bugzilla.gnome.org/show_bug.cgi?id=633028

 js/ui/panel.js |    5 ++++-
 1 files changed, 4 insertions(+), 1 deletions(-)
---
diff --git a/js/ui/panel.js b/js/ui/panel.js
index 3e58332..3196fc1 100644
--- a/js/ui/panel.js
+++ b/js/ui/panel.js
@@ -528,6 +528,9 @@ const AppMenuButton = new Lang.Class({
         if (targetApp) {
             this._appMenuNotifyId = targetApp.connect('notify::menu', Lang.bind(this, this._sync));
             this._actionGroupNotifyId = targetApp.connect('notify::action-group', Lang.bind(this, this._sync));
+        } else {
+            this._appMenuNotifyId = 0;
+            this._actionGroupNotifyId = 0;
         }
 
         this._targetApp = targetApp;
@@ -549,7 +552,7 @@ const AppMenuButton = new Lang.Class({
     _maybeSetMenu: function() {
         let menu;
 
-        if (this._targetApp.action_group) {
+        if (this._targetApp.action_group && this._targetApp.menu) {
             if (this.menu instanceof PopupMenu.RemoteMenu &&
                 this.menu.actionGroup == this._targetApp.action_group)
                 return;