我似乎很难在Java中获得多个监视器的大小;
所以我做了一个应该显示屏幕尺寸的小窗口,它适用于我的主显示器,现在我希望能够确定它所在的显示器的大小,我使用getLocation()
要知道我的JFrame在哪里,但我不知道如何获得该监视器的大小,我只能得到主要的,甚至他们的总大小。
答案 0 :(得分:2)
您需要深入GraphicsEnvironment
,这样您就可以访问系统上提供的所有GraphicsDevice
。
基本上从那里开始,你需要遍历每个GraphicsDevice
并测试窗口是否在给定GraphicsDevice
有趣的部分是,如果窗口跨越多个屏幕,该怎么办......
public static GraphicsDevice getGraphicsDevice(Component comp) {
GraphicsDevice device = null;
GraphicsEnvironment ge = GraphicsEnvironment.getLocalGraphicsEnvironment();
GraphicsDevice lstGDs[] = ge.getScreenDevices();
ArrayList<GraphicsDevice> lstDevices = new ArrayList<GraphicsDevice>(lstGDs.length);
if (comp != null && comp.isVisible()) {
Rectangle parentBounds = comp.getBounds();
/*
* If the component is not a window, we need to find its location on the
* screen...
*/
if (!(comp instanceof Window)) {
Point p = new Point(0, 0);
SwingUtilities.convertPointToScreen(p, comp);
parentBounds.setLocation(p);
}
// Get all the devices which the window intersects (ie the window might expand across multiple screens)
for (GraphicsDevice gd : lstGDs) {
GraphicsConfiguration gc = gd.getDefaultConfiguration();
Rectangle screenBounds = gc.getBounds();
if (screenBounds.intersects(parentBounds)) {
lstDevices.add(gd);
}
}
// If there is only one device listed, return it...
// Otherwise, if there is more then one device, find the device
// which the window is "mostly" on
if (lstDevices.size() == 1) {
device = lstDevices.get(0);
} else if (lstDevices.size() > 1) {
GraphicsDevice gdMost = null;
float maxArea = 0;
for (GraphicsDevice gd : lstDevices) {
int width = 0;
int height = 0;
GraphicsConfiguration gc = gd.getDefaultConfiguration();
Rectangle bounds = gc.getBounds();
Rectangle2D intBounds = bounds.createIntersection(parentBounds);
float perArea = (float) ((intBounds.getWidth() * intBounds.getHeight()) / (parentBounds.width * parentBounds.height));
if (perArea > maxArea) {
maxArea = perArea;
gdMost = gd;
}
}
if (gdMost != null) {
device = gdMost;
}
}
}
return device;
}