扩展另一个Stackoverflow问题,我已经实现了一些手势检测代码,以便我可以检测我的listview(在FrameLayout中)中的行何时被刷过.我跟着Damian的问题/答案,关于如何从适配器获取单独的行/视图.
How to get location (on screen) of row in listview
How to get location (on screen) of row in listview
我在我的onFling中有代码获取行的视图,并尝试在我的xml布局中设置为不可见的删除按钮可见.但是,这不会发生.我想知道如何在刷卡的列表视图中显示按钮？
class MyGestureDetector extends SimpleOnGestureListener {
@Override
public boolean onFling(MotionEvent e1,MotionEvent e2,float velocityX,float velocityY) {
try {
if (e2.getX() - e1.getX() > SWIPE_MIN_DISTANCE
&& Math.abs(velocityX) > SWIPE_THRESHOLD_VELOCITY) {
int itemId = MyClass.this.lv.pointToPosition(
(int) e1.getX(),(int) e1.getY());
Log.v("item id",String.valueOf(itemId));
View v = MyClass.this.adapter
.getViewOnScreen(itemId);
Button delete = (Button) v.findViewById(R.id.button_delete);
delete.setVisibility(View.VISIBLE);
//MyClass.this.adapter.notifyDataSetChanged();
}
} catch (Exception e) {
// nothing
}
return false;
}
}
我的列表适配器代码与引用的问题相同.
编辑：我尝试在列表视图上使用getChildAt()来获取行的视图,当有一个或更少的项目时,这可以工作,但是当返回错误的视图时,错误的删除按钮变为可见.
编辑2：我使用问题here上的答案让它工作：
解决方法
我曾经在我的应用程序中实现了类似的功能.我这样做的方式：
public class MyGestureDetector extends SimpleOnGestureListener {
private ListView list;
public MyGestureDetector(ListView list) {
this.list = list;
}
//CONDITIONS ARE TYPICALLY VELOCITY OR DISTANCE
@Override
public boolean onFling(MotionEvent e1,float velocityY) {
if (INSERT_CONDITIONS_HERE)
if (showDeleteButton(e1))
return true;
return super.onFling(e1,e2,velocityX,velocityY);
}
private boolean showDeleteButton(MotionEvent e1) {
int pos = list.pointToPosition((int)e1.getX(),(int)e1.getY());
return showDeleteButton(pos);
}
private boolean showDeleteButton(int pos) {
View child = list.getChildAt(pos);
if (child != null){
Button delete = (Button) child.findViewById(R.id.delete_button_id);
if (delete != null)
if (delete.getVisibility() == View.INVISIBLE)
delete.setVisibility(View.VISIBLE);
else
delete.setVisibility(View.INVISIBLE);
return true;
}
return false;
}
}
这对我有用,希望你能让它发挥作用,或者它至少会给你一些暗示.
