Android LinearLayout 触摸移动

Question

我想读取触摸位置，同时触摸被移动并释放。 但是所有这些都在 LinearLayout 类中，我不知道该怎么做。 OnTouchListener 不能像这样工作:/

Answer 1

下面是android线性布局中触摸事件的处理代码

<?xml version="1.0" encoding="utf-8"?>                                      
<LinearLayout xmlns:android="http://schemas.android.com/apk/res/android"
  android:id="@+id/l_layout"
  android:orientation="vertical"
  android:layout_width="fill_parent"
  android:layout_height="fill_parent"
>
</LinearLayout>


public class MainActivity extends Activity {

  private LinearLayout llLayout;

  @Override
  public void onCreate(Bundle savedInstanceState)
  {
     super.onCreate(savedInstanceState);
     setContentView(R.layout.main);

     llLayout = (LinearLayout)findViewById(R.id.l_layout);

     llLayout.setOnTouchListener(new View.OnTouchListener() {

    @Override
    public boolean onTouch(View v, MotionEvent event) {
     switch(event.getAction()) {
      case MotionEvent.ACTION_MOVE :
        Toast.makeText(MainActivity.this, "Touch coordinates : " +
            String.valueOf(event.getX()) + "x" + String.valueOf(event.getY()), Toast.LENGTH_SHORT).show();
            return true;

      case MotionEvent.ACTION_DOWN :
        Toast.makeText(MainActivity.this, "Touch coordinates : " +
            String.valueOf(event.getX()) + "x" + String.valueOf(event.getY()), Toast.LENGTH_SHORT).show();
            return true;
       default :
        return super.onTouch(v,event);
       }
      }
   });
 }