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Add busy property to ShellApp

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Using a separate property to show when the application is busy rather
than cramming it into the state property makes the code clearer. In most
places we only care if an app is running or not, not whether it is
actually busy.

https://bugzilla.gnome.org/show_bug.cgi?id=736492
This commit is contained in:
Phillip Wood
2014-09-16 11:08:19 +01:00
parent e00bfcc2cf
commit 546ae00854
4 changed files with 40 additions and 13 deletions
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8
js/ui/panel.js
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@ -181,6 +181,7 @@ const AppMenuButton = new Lang.Class({
this._targetApp = null;
this._appMenuNotifyId = 0;
this._actionGroupNotifyId = 0;
this._busyNotifyId = 0;
let bin = new St.Bin({ name: 'appMenu' });
bin.connect('style-changed', Lang.bind(this, this._onStyleChanged));
@ -457,12 +458,17 @@ const AppMenuButton = new Lang.Class({
this._targetApp.disconnect(this._actionGroupNotifyId);
this._actionGroupNotifyId = 0;
}
if (this._busyNotifyId) {
this._targetApp.disconnect(this._busyNotifyId);
this._busyNotifyId = 0;
}
this._targetApp = targetApp;
if (this._targetApp) {
this._appMenuNotifyId = this._targetApp.connect('notify::menu', Lang.bind(this, this._sync));
this._actionGroupNotifyId = this._targetApp.connect('notify::action-group', Lang.bind(this, this._sync));
this._busyNotifyId = this._targetApp.connect('notify::busy', Lang.bind(this, this._sync));
this._label.setText(this._targetApp.get_name());
this.actor.set_accessible_name(this._targetApp.get_name());
}
@ -476,7 +482,7 @@ const AppMenuButton = new Lang.Class({
let isBusy = (this._targetApp != null &&
(this._targetApp.get_state() == Shell.AppState.STARTING ||
this._targetApp.get_state() == Shell.AppState.BUSY));
this._targetApp.get_busy()));
if (isBusy)
this.startAnimation();
else