我试图实现一个具有以下结构的java应用程序。
我的问题是
当我从videoplayer线程调用引用线程时,视频仍会在引号表单上播放。
当我使用动作事件更改视频网址时,它只会添加当前新玩家。 恩。当我按下视频2按钮
class VideoPlayer implements Runnable,ActionListener{
private videoappMidlet MIDlet;
VideoComponent vc;
Button Videos,quotes,video1,video2,video3;
Form videoplayer;
Thread thread;
public VideoPlayer(videoappMidlet MIDlet){
this.MIDlet = MIDlet;
}
public void run(){
try{
videoplayer=new Form();
video1=new Button("video1");
.......
vc = VideoComponent.createVideoPeer("http://localhost/video1.mpg");
vc.start();
quotes.addActionListener((ActionListener) this);
........
videoplayer.addComponent(vc);
........
videoplayer.show();
}catch(Exception error){
System.err.println(error.toString());
}
}
public void start(){
thread = new Thread(this);
try{ thread.start();}
catch(Exception error){}
}
public void actionPerformed(ActionEvent ae) {
if((ae.getSource()==Quotes))
{
Quotes tp = new Quotes(this.MIDlet);
tp.start();
}
if(ae.getSource()==video1)
{
try {
vc = VideoComponent.createVideoPeer("http://localhost/video1.mpg");
vc.start();
} catch (IOException ex) {
ex.printStackTrace();
}
}
....
}
}
class Quotes implements Runnable,ActionListener {
private videoappMidlet MIDlet;
Button Videos,quotes;
Form quote;
Thread thread;
public Quotes(videoappMidlet MIDlet){
this.MIDlet = MIDlet;
}
public void run(){
try{
quote=new Form();
Videos=new Button("Videos");
........
quote.addComponent(Videos);
........
Videos.addActionListener(this);
........
quote.show();
}catch(Exception error){
System.err.println(error.toString());
}
}
public void start(){
thread = new Thread(this);
try{ thread.start();}
catch(Exception error){}
}
public void actionPerformed(ActionEvent ae) {
if(ae.getSource()==Videos)
{
VideoPlayer vp = new VideoPlayer(this.MIDlet);
vp.start();
}
}
}
public class videoappMidlet extends MIDlet implements ActionListener{
Button play,quote;
Form home;
public void startApp() {
Display.init(this);
home=new Form();
play.addActionListener(this);
quote.addActionListener(this);
home.show();
}
public void actionPerformed(ActionEvent ae) {
if(ae.getSource()==play)
{
VideoPlayer vp = new VideoPlayer(this);
vp.start();
}
if(ae.getSource()==quote)
{
Quotes tp = new Quotes(this);
tp.start();
}
}
}
答案 0 :(得分:2)
通常,JavaME中的视频不保证正在播放的图层。 LWUIT尝试无缝地暂停视频播放器,例如用户界面上的对话框。
作为附注,LWUIT不是线程安全的,您不能使用单独的线程来访问UI,因为将在不同平台上破解。