我想结合我在网上找到的两个简单的应用程序。我有一个线程和5秒后,我的传感器列表应显示Toast消息。但没有任何反应..线程不是工作我认为我搞砸了一切,请你帮忙。我真的很appriciate我的线程不工作
public class MainActivity extends Activity{
List<String>sName=new ArrayList<String>();
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
Toast.makeText(this, "Loadingg", Toast.LENGTH_LONG).show();
Thread thr=new Thread(){
@Override
public void run(){
try {
sleep(5000);
StringBuilder message=DisplaySensors();
Toast.makeText(getApplicationContext(),message, Toast.LENGTH_LONG).show();
} catch (Exception e) {
// TODO: handle exception
}
}
private StringBuilder DisplaySensors() {
SensorManager sm=(SensorManager)getSystemService(Context.SENSOR_SERVICE);
List<Sensor>sList=sm.getSensorList(Sensor.TYPE_ALL);
StringBuilder sb=new StringBuilder();
for (int i = 0; i <sList.size(); i++) {
sb.append(((Sensor)sList.get(i)).getName()).append("\n");
}
return sb;
}
};
thr.start();
}
你是一个'Thread'这是不对的内部显示'Toast'。它会崩溃吗? – Raghunandan
No Raghunandan它does not – user3468916
它不会崩溃,因为你有一个未处理的尝试捕获.. –