在当前监视器上获取组件位置

Question

我想设置JPopupMenu打开菜单的按钮的y位置的位置.我的代码在我的第一台显示器上工作正常,但在我的第二台显示器上失败,它有不同的高度.问题是getLocationOnScreen()提供相对于主屏幕的位置,而不是显示组件的实际屏幕.

我的代码:

// screenSize represents the size of the screen where the button is
// currently showing
final Rectangle screenSize = dateButton.getGraphicsConfiguration().getBounds();

final int yScreen = screenSize.height;
int preferredY;

// getLocationOnScreen does always give the relative position to the main screen
if (getLocationOnScreen().y + dateButton.getHeight() + datePopup.getPreferredSize().height > yScreen) {
  preferredY = -datePopup.getPreferredSize().height;
} else {
  preferredY = getPreferredSize().height;
}

datePopup.show(DateSpinner.this, 0, preferredY);

如何在实际监视器上获取组件的位置？

Answer 1

我使用第二个屏幕的边界得到了一个解决方案,这很简单:

public static Point getLocationOnCurrentScreen(final Component c) {
  final Point relativeLocation = c.getLocationOnScreen();

  final Rectangle currentScreenBounds = c.getGraphicsConfiguration().getBounds();

  relativeLocation.x -= currentScreenBounds.x;
  relativeLocation.y -= currentScreenBounds.y;

  return relativeLocation;
}

谢谢你的回答!