我不得不做類似的事情,發現它很混亂,因爲onExposed()可以在不確定的時間被多次調用。
爲了檢測屏幕A(主屏幕)中屏幕B的返回,我使用了屏幕B的onUiEngineAttached(false),它在彈出時被調用。
要使用回撥:
public interface Ievent {
public void backFromScreenBEvent();
}
屏幕答:
public class ScreenA extends MainScreen implements Ievent
{
private ScreenB screenB;
// constructor
public ScreenA()
{
screenB = new ScreenB(this); // pass over Ievent
// ....
}
public void backFromScreenBEvent()
{
// screen B is returning, do something
}
屏幕B:
public final class ScreenB extends MainScreen
{
private Ievent event;
// constructor
public ScreenB(final Ievent event)
{
this.event = event;
// ...
}
protected void onUiEngineAttached(boolean attached) {
super.onUiEngineAttached(attached);
if (!attached) {
event.backFromScreenBEvent(); // notify event
}
}
來源
2011-11-22 22:27:29
bob
乾草ü如何使用類Application.activate方法? – Kirti 2013-11-23 14:06:57