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popup-menu: Do not open empty menus

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There is little point in showing an empty menu, so return early from
open() if the menu does not contain any items.

https://bugzilla.gnome.org/show_bug.cgi?id=643867
This commit is contained in:
Florian Müllner 2011-11-15 14:58:55 +01:00
parent 1b64b09532
commit 55308917f9
1 changed files with 13 additions and 0 deletions
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13
js/ui/popupMenu.js
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@ -870,6 +870,10 @@ const PopupMenuBase = new Lang.Class({
return menuItem; return menuItem;
}, },
isEmpty: function() {
return this.box.get_children().length == 0;
},
isChildMenu: function(menu) { isChildMenu: function(menu) {
return this._childMenus.indexOf(menu) != -1; return this._childMenus.indexOf(menu) != -1;
}, },
@ -1195,6 +1199,9 @@ const PopupMenu = new Lang.Class({
if (this.isOpen) if (this.isOpen)
return; return;
if (this.isEmpty())
return;
this.isOpen = true; this.isOpen = true;
this._boxPointer.setPosition(this.sourceActor, this._arrowAlignment); this._boxPointer.setPosition(this.sourceActor, this._arrowAlignment);
@ -1286,6 +1293,9 @@ const PopupSubMenu = new Lang.Class({
if (this.isOpen) if (this.isOpen)
return; return;
if (this.isEmpty())
return;
this.isOpen = true; this.isOpen = true;
this.actor.show(); this.actor.show();
@ -1530,6 +1540,9 @@ const PopupComboMenu = new Lang.Class({
if (this.isOpen) if (this.isOpen)
return; return;
if (this.isEmpty())
return;
this.isOpen = true; this.isOpen = true;
let [sourceX, sourceY] = this.sourceActor.get_transformed_position(); let [sourceX, sourceY] = this.sourceActor.get_transformed_position();