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

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1 回答 1

1

下面是在android的线性布局中处理触摸事件的代码

<?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);
       }
      }
   });
 }
于 2012-10-31T23:50:52.420 回答