我有一个非常大的应用程序，它有多个对话框。我的任务是确保一个不完全可见的对话框（因为用户将其从可见屏幕区域拉出）被移回屏幕中心。
当我只处理一个屏幕时，这没问题。 它工作得很好......但是，这个应用程序的大多数用户在他们的桌面上有两个屏幕...
当我试图弄清楚对话框显示在哪个屏幕上并将其放在特定屏幕上时，......好吧，它实际上是中心，但在主屏幕上（可能不是屏幕，对话框显示上）。
为了向您展示我到目前为止的想法，这是代码......
/**
* Get the number of the screen the dialog is shown on ...
*/
private static int getActiveScreen(JDialog jd) {
int screenId = 1;
GraphicsEnvironment ge = GraphicsEnvironment.getLocalGraphicsEnvironment();
GraphicsDevice[] gd = ge.getScreenDevices();
for (int i = 0; i < gd.length; i++) {
GraphicsConfiguration gc = gd[i].getDefaultConfiguration();
Rectangle r = gc.getBounds();
if (r.contains(jd.getLocation())) {
screenId = i + 1;
}
}
return screenId;
}
/**
* Get the Dimension of the screen with the given id ...
*/
private static Dimension getScreenDimension(int screenId) {
Dimension d = new Dimension(0, 0);
if (screenId > 0) {
GraphicsEnvironment ge = GraphicsEnvironment.getLocalGraphicsEnvironment();
DisplayMode mode = ge.getScreenDevices()[screenId - 1].getDisplayMode();
d.setSize(mode.getWidth(), mode.getHeight());
}
return d;
}
/**
* Check, if Dialog can be displayed completely ...
* @return true, if dialog can be displayed completely
*/
private boolean pruefeDialogImSichtbarenBereich() {
int screenId = getActiveScreen(this);
Dimension dimOfScreen = getScreenDimension(screenId);
int xPos = this.getX();
int yPos = this.getY();
Dimension dimOfDialog = this.getSize();
if (xPos + dimOfDialog.getWidth() > dimOfScreen.getWidth() || yPos + dimOfDialog.getHeight() > dimOfScreen.getHeight()) {
return false;
}
return true;
}
/**
* Center Dialog...
*/
private void zentriereDialogAufMonitor() {
this.setLocationRelativeTo(null);
}
调试时我发现getActiveScreen()似乎没有像我那样工作的事实;它似乎总是返回2（这是一种废话，因为它意味着对话框总是显示在第二个监视器中......这当然不是事实）。
任何人都知道如何将对话框放在实际显示的屏幕上？
答案 0 :(得分：1)
您的getActiveScreen方法有效，但使用了包含窗口左上角的屏幕。如果您使用Component.getGraphicsConfiguration（），它将为您提供哪个屏幕具有窗口像素的大部分。 setLocationRelativeTo(null)在这里没有帮助，因为它总是使用主屏幕。以下是解决方法：
static boolean windowFitsOnScreen(Window w) {
return w.getGraphicsConfiguration().getBounds().contains(w.getBounds());
}
static void centerWindowToScreen(Window w) {
Rectangle screen = w.getGraphicsConfiguration().getBounds();
w.setLocation(
screen.x + (screen.width - w.getWidth()) / 2,
screen.y + (screen.height - w.getHeight()) / 2
);
}
然后你可以这样做：
JDialog jd;
...
if (!windowFitsOnScreen(jd)) centerWindowToScreen(jd);
将对话框置于最近的屏幕（监视器）中心。您可能需要确保最初显示/定位对话框。
答案 1 :(得分：1)
我不确定这将有多大用处，但这是我在尝试确定Windows图形设备时使用的代码。
我作弊一点，我倾向于使用Component并允许实用程序方法找到顶级窗口或使用Component的屏幕点。
/**
* Returns the GraphicsDevice that the specified component appears the most on.
*/
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);
}
for (GraphicsDevice gd : lstGDs) {
GraphicsConfiguration gc = gd.getDefaultConfiguration();
Rectangle screenBounds = gc.getBounds();
if (screenBounds.intersects(parentBounds)) {
lstDevices.add(gd);
}
}
if (lstDevices.size() == 1) {
device = lstDevices.get(0);
} else {
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;
}
/**
* Returns the GraphicsDevice at the specified point
*/
public static GraphicsDevice getGraphicsDeviceAt(Point pos) {
GraphicsDevice device = null;
GraphicsEnvironment ge = GraphicsEnvironment.getLocalGraphicsEnvironment();
GraphicsDevice lstGDs[] = ge.getScreenDevices();
List<GraphicsDevice> lstDevices = new ArrayList<GraphicsDevice>(lstGDs.length);
for (GraphicsDevice gd : lstGDs) {
GraphicsConfiguration gc = gd.getDefaultConfiguration();
Rectangle screenBounds = gc.getBounds();
if (screenBounds.contains(pos)) {
lstDevices.add(gd);
}
}
if (lstDevices.size() > 0) {
device = lstDevices.get(0);
}
return device;
}
/**
* Returns the Point that would allow the supplied Window to be
* centered on it's current graphics device.
*
* It's VERY important that the Window be seeded with a location
* before calling this method, otherwise it will appear on the
* device at 0x0
*
* @param window
* @return
*/
public static Point centerOfScreen(Window window) {
// Try and figure out which window we actually reside on...
GraphicsDevice gd = getGraphicsDeviceAt(window.getLocation());
GraphicsConfiguration gc = gd.getDefaultConfiguration();
Insets screenInsets = Toolkit.getDefaultToolkit().getScreenInsets(gd.getDefaultConfiguration());
Rectangle bounds = gc.getBounds();
Dimension size = bounds.getSize();
size.width -= (screenInsets.left + screenInsets.right);
size.height -= (screenInsets.top + screenInsets.bottom);
int width = window.getWidth();
int height = window.getHeight();
int xPos = screenInsets.left + ((size.width - width) / 2);
int yPos = screenInsets.top + ((size.height - height) / 2);
return new Point(xPos, yPos);
}
答案 2 :(得分：0)
这里是用于居中窗口位置的代码。
//Center the window
Dimension screenSize = Toolkit.getDefaultToolkit().getScreenSize();
Dimension frameSize = frame.getSize();
if (frameSize.height > screenSize.height) {
frameSize.height = screenSize.height;
}
if (frameSize.width > screenSize.width) {
frameSize.width = screenSize.width;
}
frame.setLocation((screenSize.width - frameSize.width) / 2, (screenSize.height - frameSize.height) / 2);
使用frame您也可以使用该对话框。