win: Force menuWillShow to be called when setting window menu.

This commit is contained in:
Cheng Zhao 2014-05-26 09:38:04 +08:00
parent 0e5f697b48
commit 6ee437e9bf
2 changed files with 11 additions and 7 deletions

View file

@ -230,7 +230,7 @@ void Menu::BuildPrototype(v8::Isolate* isolate,
.SetMethod("isEnabledAt", &Menu::IsEnabledAt)
.SetMethod("isVisibleAt", &Menu::IsVisibleAt)
#if defined(OS_WIN) || defined(TOOLKIT_GTK)
.SetMethod("attachToWindow", &Menu::AttachToWindow)
.SetMethod("_attachToWindow", &Menu::AttachToWindow)
#endif
.SetMethod("_popup", &Menu::Popup);
}

View file

@ -91,18 +91,22 @@ Menu::insert = (pos, item) ->
@items.splice pos, 0, item
@commandsMap[item.commandId] = item
Menu::attachToWindow = (window) ->
@_callMenuWillShow() if process.platform is 'win32'
@_attachToWindow window
# Force menuWillShow to be called
Menu::_callMenuWillShow = ->
@delegate?.menuWillShow()
item.submenu._callMenuWillShow() for item in @items when item.submenu?
applicationMenu = null
Menu.setApplicationMenu = (menu) ->
throw new TypeError('Invalid menu') unless menu?.constructor is Menu
applicationMenu = menu # Keep a reference.
if process.platform is 'darwin'
# Force menuWillShow to be called
menuWillShow = (menu) ->
menu.delegate?.menuWillShow()
menuWillShow item.submenu for item in menu.items when item.submenu?
menuWillShow menu
menu._callMenuWillShow()
bindings.setApplicationMenu menu
else
windows = BrowserWindow.getAllWindows()