在android中编写线程和处理程序?

时间:2011-02-13 17:21:04

标签: android multithreading handler

我执行了以下代码

public class Activity_Threads_Handler extends Activity {


private int mSec;

TextView tv;

protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.main);
    tv = (TextView) findViewById(R.id.TextView01);
    startHeavyDutyStuff();
}

void startHeavyDutyStuff() {

    // Here is the heavy-duty thread
    Thread t = new Thread() {

        public void run() {
            try {
                while (true) {

                    mSec = Calendar.getInstance().get(Calendar.MINUTE);
                    // Send update to the main thread
                    messageHandler.sendMessage(Message.obtain(
                            messageHandler, mSec));

                    Thread.sleep(1000);
                }
            } catch (InterruptedException e) {
                e.printStackTrace();
            }

        }
    };
    t.start();
}

// Instantiating the Handler associated with the main thread.
private Handler messageHandler = new Handler() {

    @Override
    public void handleMessage(Message msg) {
        tv.setText(msg.what);
    }

};

}

上面的代码是显示当前第二个到TextView不断更新它...

我的logcat

中出现以下错误
  

致命的例外:主要   android.content.res.Resources $ NotFoundException:字符串资源ID#x22   在android.content.res.Resources.getText(Resources.java:201)   在android.widget.TextView.setText(TextView.java:2857)   at com.kpj4s.Threads_Handler.Activity_Threads_Handler $ 1.handleMessage(Activity_Threads_Handler.java:56)   在android.os.Handler.dispatchMessage(Handler.java:99)   在android.os.Looper.loop(Looper.java:123)   在android.app.ActivityThread.main(ActivityThread.java:3647)   at java.lang.reflect.Method.invokeNative(Native Method)   在java.lang.reflect.Method.invoke(Method.java:507)   在com.android.internal.os.ZygoteInit $ MethodAndArgsCaller.run(ZygoteInit.java:839)   在com.android.internal.os.ZygoteInit.main(ZygoteInit.java:597)   在dalvik.system.NativeStart.main(本地方法)

有人可以帮我解决.. 谢谢:))

2 个答案:

答案 0 :(得分:2)

问题是:

tv.setText(msg.what);

你想做什么?

问题是msg.what是一个整数,setText接收:String或整数是一个android资源(例如R.string.something)。因此,如果您想要显示msg.what内容,请使用tv.setText(String.valueOf(msg.what))

答案 1 :(得分:1)

TextView.setText(int)期望整数参数是资源标识符。您需要将其转换为String,即您想要:

tv.setText(Integer.toString(msg.what));