Android的：如何拖动（移动）PopupWindow？ [英] Android: How to drag(move) PopupWindow?

问题描述

我希望能够移动PopupWindow触摸拖动。 我不希望用户界面更新在触摸释放。我想PopupWindow按照我联系。

I want to be able to move PopupWindow on touch dragging. I don't want UI to update on the release of the touch. I want PopupWindow to follow my touch.

这是一些我在做什么：

mView = mLayoutInflater.inflate(R.layout.popup,
                null);
mPopupWindow = new PopupWindow(mView,
               LayoutParams.WRAP_CONTENT, LayoutParams.WRAP_CONTENT, false);
mPopupWindow.showAtLocation(parentView, Gravity.CENTER, -5, 30);

mView.setOnTouchListener(new OnTouchListener() {
        private int dx = 0;
        private int dy = 0;

        @Override
        public boolean onTouch(View view, MotionEvent motionEvent) {
            switch (motionEvent.getAction()) {
                case MotionEvent.ACTION_DOWN:
                    dx = (int) motionEvent.getX();
                    dy = (int) motionEvent.getY();
                    break;
                case MotionEvent.ACTION_MOVE:
                    int x = (int) motionEvent.getX();
                    int y = (int) motionEvent.getY();
                    int left =  (x - dx);
                    int top =   (y - dy);
                    Log.d("test", "x: " + left + " y: " + top);
                    mPopupWindow.update(left, top, -1, -1);
                    break;
            }
            return true;
        }
    });

会发生什么事，我拖了弹出式窗口，它闪烁来回在原来的位置，并在我的手指。

What happens is, as I drag the popup window, it flickers back and forth on the original location and where my finger is.

闪烁logcat的结果：

Flickering Logcat Result:

x: -44 y: 4
x: -43 y: 37
x: -46 y: 4
x: -46 y: 38
x: -48 y: 4
x: -47 y: 38
x: -50 y: 4

但是，如果我删除（注释掉）mPopupWindow.update（左，上，-1，-1）;，它返回正确的结果。 （但显然UI将不会更新）：

But if I remove (comment out) "mPopupWindow.update(left, top, -1, -1);", it returns correct result. (But obviously UI won't update):

x: -33 y: 0
x: -37 y: 0
x: -41 y: 0
x: -43 y: 3
x: -46 y: 3
x: -50 y: 3
x: -54 y: 4
x: -57 y: 4

我将如何正确地移动PopupWindow？

How would I move a PopupWindow correctly?

推荐答案

在Eclipse中，它说：信息getX（INT）的第一个指针指数。我假定这应该是只用于设置第一个x和y ACTION.DOWN时。当使用ACTION.MOVE，getRawX（）和getRawY（），蚀表示：在屏幕上的事件的原始位置。使用这种方式，弹出窗口将不会闪烁和移动，直到它被解雇后，将保持在它的位置。

In eclipse, when looking at the info on getX(), it says: getX(int) for the first pointer index . I am assuming that this should be the only used when setting the first x and y for ACTION.DOWN. When using ACTION.MOVE, getRawX() and getRawY(), eclipse says: original location of the event on the screen. Using it this way, popup window will not flicker and will remain at its location after moving until it's dismissed.

更新code：

case MotionEvent.ACTION_DOWN:
    dx = (int) event.getX();
    dy = (int) event.getY();
    break;

case MotionEvent.ACTION_MOVE:
    xp = (int) event.getRawX();
    yp = (int) event.getRawY();
    sides = (xp - dx);
    topBot = (yp - dy);
    popup.update(sides, topBot, -1, -1, true);
    break;

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