android退出一个activity时如何同时退出其中开启的线程

Question

新人初学android
发现在退出页面后该页面中开启的线程仍在不停的获取服务器数据
求帮忙解决问题，希望最好有个Demo可以看看
代码如下

public class WaterActivity10 extends Activity {
    private final int HIGH = 0;

    private  LineChart mChart1=null;

    private TextView Servera1=null;

    private  LineChart mChart2=null;

    private TextView Servera2=null;


    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_water10);

        Servera1=(TextView) findViewById(R.id.Servera1);
        Servera2=(TextView) findViewById(R.id.Servera2);



        //refreshChart();

        new Thread(){
            public void run(){
                try {
                    while (true) {
                        Thread.sleep(900);
                        new AnotherTask().execute("JSON");
                        Log.d("Thread1", "Thread one cnt: ");

                    }

                }
                catch (Exception  e){
                    e.printStackTrace();

                }

            }
        }.start();





    }


    private class AnotherTask extends AsyncTask<String, Void, String> {
        @Override
        protected void onPostExecute(String result) {
            //对UI组件的更新操作
            addEntry(mChart1);
            addEntry(mChart2);
        }
        @Override
        protected String doInBackground(String... params) {
            //耗时的操作
            return params[0];
        }
    }

    private void addEntry(LineChart mChart) {

        LineData data = mChart.getData();

        data.addXValue((data.getXValCount()) + "");

        MyApplication app=(MyApplication)getApplication();


        int str1= Tools.getnumber(app.getServerAddress());
        int str2= Tools.getnumber(app.getServerAddress());


        Servera1.setText(String.valueOf(str1)+"号车");
        Servera2.setText(String.valueOf(str2)+"t");


    }

}

Answer 1

这种写法不太推荐，提供一个解决你说的问题：设置线程的执行条件，在onDestroy中将条件置成不可执行。

  while (true) {}