android - 如何获得拖动的释放 X 和 Y 位置？

Question

我有一个 DragListener，我想从中获取 View 释放位置的坐标。但无论我把它放在哪里，我都会得到 x = 0 和 y = 0。显然我做错了什么。如何获取发布点的x和y？

this.setOnDragListener(new View.OnDragListener() {
    CustomIcon temp;
    CustomIcon dragging;
    Boolean thisOne;

    public boolean onDrag(View v, DragEvent event) {

        for(int i= 0; i<mIcons.size();i++) {
            temp = mIcons.get(i);
            thisOne = temp.getDragging();
            if(thisOne) {
                dragging = temp;
            }
        }
        final int action = event.getAction();
        switch (action) {
            case DragEvent.ACTION_DRAG_STARTED: {

                Toast.makeText(getContext(), "Dragging: "+v+" Width:"+dragging.getWidth()+ " X:"+dragging.getX(),Toast.LENGTH_SHORT).show();        
                dragging.setVisibility(View.INVISIBLE);

            } break;

            case DragEvent.ACTION_DRAG_ENDED: {

                dragging.setVisibility(View.VISIBLE);
                dragging.stopDragging();

                // WHERE MY PROBLEM IS:
                Toast.makeText(getContext(), "Result: "+event.getResult()+" endX:"+(int) event.getX() + " endY:"+event.getY(),Toast.LENGTH_SHORT).show();
                final boolean dropped = event.getResult();

            } break;
        }
        return false;
    }
});

Answer 1

我认为你的案例听错了事件。 http://developer.android.com/reference/android/view/DragEvent.html#getX ()

public float getX ()

Gets the X coordinate of the drag point. The value is only valid if the event action is ACTION_DRAG_LOCATION or ACTION_DROP.

尝试将您的案例更改为 DragEvent.ACTION_DROP