Question

android popupwindow怎么合理控制弹出位置

Answer 1

/**
* 计算出来的位置，y方向就在anchorView的上面和下面对齐显示，x方向就是与屏幕右边对齐显示
* 如果anchorView的位置有变化，就可以适当自己额外加入偏移来修正
* @param anchorView 呼出window的view
* @param contentView window的内容布局
* @return window显示的左上角的xOff,yOff坐标
*/
private static int[] calculatePopWindowPos(final View anchorView, final View contentView) {
final int windowPos[] = new int[2];
final int anchorLoc[] = new int[2];
　　　　 // 获取锚点View在屏幕上的左上角坐标位置
anchorView.getLocationOnScreen(anchorLoc);
final int anchorHeight = anchorView.getHeight();
// 获取屏幕的高宽
final int screenHeight = ScreenUtils.getScreenHeight(anchorView.getContext());
final int screenWidth = ScreenUtils.getScreenWidth(anchorView.getContext());
contentView.measure(View.MeasureSpec.UNSPECIFIED, View.MeasureSpec.UNSPECIFIED);
// 计算contentView的高宽
final int windowHeight = contentView.getMeasuredHeight();
final int windowWidth = contentView.getMeasuredWidth();
// 判断需要向上弹出还是向下弹出显示
final boolean isNeedShowUp = (screenHeight - anchorLoc[1] - anchorHeight < windowHeight);
if (isNeedShowUp) {
windowPos[0] = screenWidth - windowWidth;
windowPos[1] = anchorLoc[1] - windowHeight;
} else {
windowPos[0] = screenWidth - windowWidth;
windowPos[1] = anchorLoc[1] + anchorHeight;
}
return windowPos;
}