First of all, I'd like to mention that I found that related post How to get the mouse position on the screen in Qt? but it "just didn't work" for me. I made some tests, and the results didn't work as I expected, so I decided to make a new post to talk about the test I made and to find an alternative solution.
That's the code I used to make the test:
QScreen *screen0 = QApplication::screens().at(0);
QScreen *screen1 = QApplication::screens().at(1);
printf("screen0 %s \n", screen0->name().toStdString().c_str());
printf("screen1 %s \n", screen1->name().toStdString().c_str());
// Position on first screen.
QPoint pos0 = QCursor::pos(screen0);
// Position on second screen.
QPoint pos1 = QCursor::pos(screen1);
printf("pos 0: %d, %d \n", pos0.x(), pos0.y());
printf("pos 1: %d, %d \n", pos1.x(), pos1.y());
// Get position without screen.
QPoint pos = QCursor::pos();
printf("pos: %d, %d \n", pos.x(), pos.y());
What I was expecting, is that only one screen would return a valid position, since the cursor is only at one screen, not on both. But it's not the case, the both positions (pos0 and pos1) has the exactly same value, as we can see on the output:
screen0 DVI-D-0
screen1 HDMI-0
pos 0: 1904, 1178
pos 1: 1904, 1178
pos: 1904, 1178
Since the both positions has the same values, I can't know at which screen is the cursor. I don't know if that's a normal behavior or a bug, since the documentation doesn't say what happens when the screen argument isn't the screen where the mouse is.
My idea, is to open/launch an application (executed by a Qt daemon that must detect the selected screen) to the screen where the mouse is. I know that with libX11 it's possible, because I did it in the past, but I need to work with Qt 5, and I can't figure out how to do detect the selected screen with Qt.
I also made other tests, using QApplication and QDesktopWidget classes with no luck.
That's really weird. As a workaround, you could try this:
QPoint globalCursorPos = QCursor::pos();
int mouseScreen = qApp->desktop()->screenNumber(globalCursorPos);
Now you know which screen the cursor is in. Then you could find the cursor position within that screen doing this:
QRect mouseScreenGeometry = qApp->desktop()->screen(mouseScreen)->geometry();
QPoint localCursorPos = globalCursorPos - mouseScreenGeometry.topLeft();
This may seem like a trivial solution, but on my KDE it works (I ran into the same problems originally).
If you want to determine the local mouse coordinates with respect to a widget (this will be in device pixels and relative to the top left corner of the widget I believe) you can use
QWidget::mapFromGlobal(QCursor::pos());
i.e. call this->mapFromGlobal.
To figure out on which screen you are, you can iterate throught QGuiApplication::screens() and check whether the cursor fits in the geometry of the screen.
Here is a more complex example to compute the native cursor position (note the additional work needed to work with High DPI screens):
QPoint getNativeCursorPosition()
{
QPoint pos = cursorPosToNative(QCursor::pos());
// Cursor positions from Qt are calculated in a strange way, the offset to
// the origin of the current screen is in device-independent pixels while
// the origin itself is native!
for (QScreen *screen : QGuiApplication::screens()) {
QRect screenRect = screen->geometry();
if (screenRect.contains(pos)) {
QPoint origin = screenRect.topLeft();
return origin + (pos - origin) * screen->devicePixelRatio();
}
}
// should not happen, but try to find a good fallback.
return pos * qApp->devicePixelRatio();
}
This may work for you?
It did for me
QDesktopWidget *widget = QApplication::desktop();
QPosition globalCursorPosition = widget->cursor().pos();
Sice it seems that it can't be done with Qt (at least with my system configuration, and it seems that also in Windows) I decided to use the libX11 to make that implementation, which works like charm.
It's not an ideal solution because I wanted to only use Qt, but it works.
With QML you can use the properties of the Screen QML Type:
Screen.virtualX : The x coordinate of the screen within the virtual desktop.
Screen.virtualY : The y coordinate of the screen within the virtual desktop.
import QtQuick 2.6
import QtQuick.Window 2.2
console.log("Pos x : " + Screen.virtualX )
console.log("Pos y : " + Screen.virtualY )
This work with single screen as well multi-monitor systems.
I recently ran into a similar problem on Qt 5.15 + Windows + mixed DPI and needed this work around within a QWindow object.
QScreen* primaryScreen = QGuiApplication::primaryScreen();
QScreen* thisScreen = screen();
qreal primaryDPR = primaryScreen->devicePixelRatio();
qreal thisDPR = thisScreen->devicePixelRatio();
qreal scale = thisDPR / primaryDPR;
QPoint pos = scale * QCursor::pos();
I'm unsure if this works on other platforms.
Related
I am working on a MFC app which is a MDI. One of the child frame uses OpenGL(mixed with fixed function and modern version) called 3d view and another child frame uses GDI called plan view. Both of the views use the same doc.
The 3d view has a function to detect if the mouse cursor is over rendered 3d model by reading pixels and check its depth value.
The function is used for WM_MOUSEMOVE and WM_LBUTTONDOWN events. Most time it works pretty well. But it failed when I move my cursor from the plan view(currently active) to the 3d view and left mouse click. The depth values read from the pixels(called from onLButtonDown) are always all zeros though it is over a model. There is no OpenGL error reported. It only fails on the first mouse click when the 3d view is not activated. Afterwards, everything works well again.
The issue doesn't happen on all machines. And it happens to me but not to another guy with the same hardware machine with me. Is that possible hardware related or a code bug?
Things tried:
I tried to increase the pixel block size much bigger but depths are still all zero.
If I click on the title bar of the 3d view to activate it first, then works.
I tried to set the 3d view active and foreground in the onLButtonDown method before reading pixels. But still failed.(btw, the 3d view should be active already before the OnLButtonDown handler via other message handler fired by the left button down).
I tried to invalidate rect before reading pixels, failed too.
The code is as below:
BOOL CMy3DView::IsOverModel(int x0, int y0, int &xM, int &yM, GLfloat &zWin, int width0 , int height0 )
{
int width = max(1,width0);
int height= max(1,height0);
CRect RectView;
GetClientRect(&RectView);
GLint realy = RectView.Height() - 1 - (GLint)y0 ; /* OpenGL y coordinate position */
std::vector<GLfloat> z(width*height);
//Read the window z co-ordinates the z value of the points in a rectangle of width*height )
xM = max(0, x0-(width-1)/2);
yM = max(0, realy-(height-1)/2);
glReadPixels(xM, yM, (GLsizei)width, (GLsizei)height, GL_DEPTH_COMPONENT, GL_FLOAT, &z[0]); OutputGlError(_T("glReadPixels")) ;
/* check pixels along concentric, nested boxes around the central point */
for (int k=0; k<=(min(height,width)-1)/2; ++k){
for (int i=-k;i<=k;++i){
xM = x0+i;
for (int j=-k;j<=k;++j){
if (abs(i)==k || abs(j)==k) {
yM = realy+j;
zWin=z[(i+(width-1)/2)+width*(j+(height-1)/2)];
if (zWin<1.0-FLT_EPSILON) break;
}
}
if (zWin<1.0-FLT_EPSILON) break;
}
if (zWin<1.0-FLT_EPSILON) break;
}
yM = RectView.Height() - 1 - yM;
if (zWin>1.0-FLT_EPSILON || zWin<FLT_EPSILON) {// z is the depth, between 0 and 1, i.e. between Near and Far plans.
xM=x0; yM=y0;
return FALSE;
}
return TRUE;
}
Just found a solution for that: I called render(GetDC) before any processing in OnLButtonDown. somehow it fixed the issue though I don't think it's necessary.
InvalideRect wont fix the issue since it will update the view for the next WM_PAINT.
Weird, since it works for some machines without the fix. Still curious about the reason.
I am currently working on a project that I use C++, OpenGL, Qt 5.9.2 and Microsoft Visual Studio Professional 2015 on a 64 bit Operating System, Windows 10 Pro.
I have a user interface that I have created and in that user interface, there is a QGLWidget, that I am using for draw processes, with other widgets like push buttons and a dockWidget. I have a class Ball and it has variables(distance(double) and angle(int)) that determines where an instance of that Ball is going to be drawn inside the QGLWidget. Ball class has got 2 more variables that is, id(int), model(String) and year(int) Every other draw process draws lines except Ball drawings.
Drawings are 2 dimensional.
Every Ball has the same color(rgb)!
First problem: I want to left click to one of the Ball instances and I want to display it's id, model and year at The dockWidget.
Second Problem: While doing the stuff that I have mentioned at the First Problem section. I want the cursor image to change while hovering above any of the Ball instances, and change back to default Windows mouse cursor while not.
I have created a function that checks if the MouseEvent is LeftClick:
void DisplayManager::mousePressEvent(QMouseEvent* ev) {
if (ev->buttons() & Qt::LeftButton) { // Balls Are Green
if(// CHECK IF THERE IS A BALL AT THE CLICKED COORDINATES) {
// DISPLAY THE X and Y OF THE BALL AT THE DOCK WIDGET
}
}
}
This is my initializeGL function: (DisplayManager is the name of my QGLWidget)
void DisplayManager::initializeGL() {
glEnable(GL_COLOR_MATERIAL); // Enables the changing of the draw color with glColor() functions
glColor3f(0.0, 1.0, 0.0);
glEnable(GL_DEPTH_TEST);
glClearColor(0, 0, 0, 1); //sets a black background 1 0 0 1
}
On the basis this is a Picking problem and there are several information about it at the internet but I am not using GLUT and I am not using any shader. So in the light of all these I was unable to find any effective solution or clue about how can I accomplish all that I want.
I would be really glad if someone could help me with at least one of these problems.
I have currently finished working with the project. I thought that I should provide an answer to my question in case someone with a similar problem comes across with my question in the future.
void DisplayManager::mousePressEvent(QMouseEvent* ev) {
// The processes that are being executed after a left mouse button click on the DisplayManager.
if (ev->buttons() & Qt::LeftButton) {
double aspectRatio = openGLWidgetWidth / openGLWidgetHeight;
int xOfClicked = ev->x() - (openGLWidgetWidth / 2);
int yOfClicked = - (ev->y() - (openGLWidgetHeight / 2));
// The variable to be used for calculating fault tolerance of the click event.
int distanceBetweenPressAndDraw = 0;
// Executes the processes inside for every ball in vector.
for (int i = 0; i < ballVector.length(); i++) {
// Calculates the screen coordinates of the i'th ball.
int rangeOfBallInScreenDistance = rangeToScreenDistance(ballVector.at(i).range);
double screenXOfBall = rangeOfBallInScreenDistance * sin(ballVector.at(i).degree * DEGTORAD);
double screenYOfBall = rangeOfBallInScreenDistance * cos(ballVector.at(i).degree * DEGTORAD);
// Calculates the distance between pressed position and the i'th ball according to the screen coordinates.
distanceBetweenPressAndDraw = sqrt(pow((screenXOfBall - xOfClicked), 2) + pow((screenYOfBall - yOfClicked), 2));
// Decides if the clicked position is a ball (considering the fault tolerance).
if (distanceBetweenPressAndDraw < 10) {
emit printXY(QPointF(xOfClicked, yOfClicked)); // Prints the screen coordinates of the clicked positions (At the Dock Widget inside Main Window).
}
}
}
}
This was the solution for my First Problem. I would be glad though if someone could answer my Second problem in a comment or answer somehow.
We are developing a macOS application whose GUI is relying on Qt.
At startup, we want to show() the QMainWindow at a specific location on the currently active screen (with multi screen systems in mind).
Is there a way to get the QScreen representing the currently active screen?
From our test, QGuiApplication::primaryScreen() is the first screen (which is consistent with the name), but we cannot find an equivalent for the active screen.
Qt5 provides functionality to do so, the QWindow::setScreen method sets the screen on which the window should be shown.
Any widget provides access to this pointer via QWidget::windowHandle():
QWidget * widget = new QWidget();
auto screens = qApp->screens();
// compute the index
widget->windowHandle()->setScreen(screens[index]);
widget->showFullScreen();
To get the screen number, you can use the mouse position and assume that the screen with the mouse is the one with the current focus:
QPoint globalCursorPos = QCursor::pos();
int mouseScreen = qApp->desktop()->screenNumber(globalCursorPos);
So the final code can be something like that:
QWidget * widget = new QWidget();
const auto globalCursorPos = QCursor::pos();
const auto mouseScreen = qApp->desktop()->screenNumber(globalCursorPos);
widget->windowHandle()->setScreen(qApp->screens()[mouseScreen]);
widget->showFullScreen();
Windows
If this approach does not fit your needs, you will need to perform some OS calls.
For instance, on Windows, you can use MonitorFromWindow:
HMONITOR active_monitor_number = MonitorFromWindow(GetActiveWindow(), MONITOR_DEFAULTTONEAREST);
If you need more information about the screen, you can use Qt or GetMonitorInfo.
I am not a Mac OS X developer, but it may exist a similar API
I did it in the following way:
#if (QT_VERSION >= QT_VERSION_CHECK(5, 10, 0))
QPoint topLeft = QPoint(0, 0);
QScreen* currentScreen = QGuiApplication::screenAt(QCursor::pos());
if (nullptr != currentScreen) {
topLeft = currentScreen->geometry().topLeft();
}
#else
QPoint topLeft =
qApp->desktop()
->screenGeometry(qApp->desktop()->screenNumber(QCursor::pos()))
.topLeft();
#endif
someWidget->move(mapFromGlobal(topLeft) + QPoint(offset, offset));
Note that, sometimes, you may get nullptr for currentScreen (in my case if primary screen is at the bottom and mouse pos at the bottom or left edge on the primary screen).
The title says it pretty much all:
I have two screens, and each time I create a QDialog it appears in the same screen as its parent.
How can I make it appear in a different screen? Or should I use a different type of top-level widget?
The code I use to create the dialog is:
QDialog my_dialog = new QDialog(this,
Qt::WindowMaximizeButtonHint |
Qt::WindowCloseButtonHint);
...
EDIT:
I have also tried using the QDesktopWidget which gives me a QScreen object that refers to the second screen. But then I don't find how to instruct the QDialog to use that QScreen (setting it as the parent doesn't work).
It is bad, that you edit your question without reading comments :(
// Your screen geometry:
QRect buildScreenGeometry()
{
auto desktop = QApplication::desktop();
QRect virtualRect;
const auto n = desktop->screenCount();
for ( auto i = 0; i < n; i++ )
virtualRect |= desktop->screenGeometry(i);
return virtualRect;
}
// Moving
auto dlg = new QDialog( someParent );
auto newPoint = QPoint( 2000, 0 ); // point on another screen
auto realPos = someParent->mapFromGlobal( newPoint );
dlg->move( realPos );
That's all.
UPDATE:
You should understand, that there are only ONE screen area with COMMON coordinate system, that contains ALL screens.
For example, you have 2 monitors with 800x600 resolution. First (main) monitor is standing left, and second standing right. In this case, coordinate system, that is available for your application is 1600x600. So, if your widget has 100x100 top-left position, on a first monitor, and you want to move it to another, you should call move(900x100); // 900 == screen1.width() + dialog.pos().x(). Then your widget will have 100x100 position on second monitor.
You should read Qt documentation.
You can use move on your QDialog, but be aware that move will set the QDialog position relative to it's parent.
You can get the Main Window's screen position and use that to setup the QDialog position. Just know that you're not guaranteed to have 2 screens on the end user machine.
For more information on move see: http://doc.qt.io/qt-5/application-windows.html#window-geometry
I try to draw a round rectangle with drawRoundedRect method directly in a QPixmap (no render engine involve here exept pure Qt one ...), I double check the size of the rectangle versus the size of my pixmap :
Pixmap : QSize(50, 73)
Rectangle: QRect(0,0 48x11)
See plenty of space ...
EDIT: some code
pixmap = QPixmap(50,73); //example size that match my case
QRectF rect(0,0,48,11);
QPainter painter(&pixmap);
painter.setRenderHint(QPainter::TextAntialiasing);
painter.setWorldMatrixEnabled(false);
painter.setPen(QPen()); //no pen
painter.setBrush(QBrush(color));
painter.drawRoundedRect(rect, 2.0, 2.0);
I disabled world transformation ...
I set set transformation to unity ...
I tried several radius (1.0,2.0,3.0,4.0) ...
I change pen width, brush color ...
But it always ends with a rectamgle with 4 diferent corners ! Like that :
I directly ouptut the pixmap to a file to be sure I wasn't scraping it during the display ... same shape.
Anyone know about Qt round rectangle with small radius ? I saw somthing about it a long time ago but I don't remenber how to deal with it !
It looks like you're not using anti-aliasing (i.e. the QPainter::Antialiasing render hint). This is a Qt quirk that occurs without it. From what I've seen/heard, the Qt devs aren't terribly concerned with fixing this (most people want anti-aliasing anyway).
The work-around (besides just using anti-aliasing) is to draw the rect yourself with QPainter::drawLine() and QPainter::drawArc(). You might have to play with numbers until it looks right -- straight calculations tend to come out a pixel or two off. Also, you might find that even with this method the lower right corner is never exactly the same as the other corners.
If you're feeling mildly ambitious, you could try fixing this and submitting a patch to Qt.
Update: Arc drawing results changed in Qt 5. In my experience, it's a big improvement.
I know this is an old problem but for Qt5 users calling setRenderHint(QPainter::Qt4CompatiblePainting); on the QPainter seems to solve the problem.
Edit:
I found a solution for generating a perfect rounded rectangle together with border color and it looks the same as the rounded rectangles used by QPushButton's border for example. This is how I implemented the paintEvent to achieve this:
void MyButtonGroup::paintEvent(QPaintEvent * e)
{
int borderSize = 5;
QColor borderColor = Qt::red;
QColor backgroundColor = Qt::blue;
int borderRadius = 3;
QPen pen;
pen.setWidth(borderSize);
pen.setColor(borderColor);
QPainter painter(this);
painter.setRenderHint(QPainter::Antialiasing);
painter.setPen(pen);
QRectF rect(rect().x() + borderSize / 2,
rect().y() + borderSize / 2,
rect().width() - borderSize,
rect().height() - borderSize);
if(borderSize % 2 == 0)
{
painter.drawRoundedRect(rect,
borderSize,
borderSize);
}
else
{
painter.drawRoundedRect(rect.translated(0.5, 0.5),
borderRadius,
borderRadius);
}
QBrush brush(backgroundColor);
pen.setBrush(brush);
painter.setBrush(brush);
if(borderSize % 2 == 0)
{
painter.drawRoundedRect(rect,
borderRadius,
borderRadius);
}
else
{
painter.drawRoundedRect(rect.translated(0.5, 0.5),
borderRadius,
borderRadius);
}
QWidget::paintEvent(e);
}
I'm posting this because I found it a bit hard to achieve this result:
Try adding half a pixel offset (e.g.: rect.translated(0.5,0.5) ):
QRectF rect(0,0,48,11);
painter.setRenderHint(QPainter::Antialiasing,false);
painter.drawRoundedRect( rect.translated(0.5,0.5), 2.0, 2.0 );
I suppose this has to do with the coordinate system placing an integer value between two pixels.
If you draw with antialiasing and use a pen of 1 pixel width then drawing at exact integer coordinates results in lines of 2 pixel width instead.
Only with this 0.5 pixel offset you'll get lines that are exactly 1 pixel wide.
QRectF rect(0,0,48,11);
painter.setRenderHint(QPainter::Antialiasing,true);
painter.setBrush(Qt::NoBrush);
painter.setPen( Qt::white );
painter.drawRoundedRect( rect.translated(0.5,0.5), 2.0,2.0 );
Best way do draw RoundRect is Path.
http://developer.nokia.com/community/wiki/Qt_rounded_rect_widget
void fillRoundRect(QPainter& painter, QRect r, int radius)
{
painter.setRenderHint(QPainter::Antialiasing,true);
QPainterPath rounded_rect;
rounded_rect.addRoundRect(r, radius, radius);
painter.setClipPath(rounded_rect);
painter.fillPath(rounded_rect,painter.brush());
painter.drawPath(rounded_rect);
}
try to play with render hints
1) disable antiAliasing;
2) enable SmoothPixmapTransform
but still no guarantee that it will help.
I have tried all tips from answers here but nothing works for me. But based on these code snippets I have found following solution:
As default set m_pPainter->setRenderHint(QPainter::Qt4CompatiblePainting, true) and only for rounded rectangles with width%2==0 disable it.
QRect rect = ConvertRectangle(rectangle);
int nPenWidth = m_pPainter->pen().width();
if ( nPenWidth % 2 == 0 )
m_pPainter->setRenderHint(QPainter::Qt4CompatiblePainting, false);
m_pPainter->drawRoundedRect(rect, dbRadiusX, dbRadiusY);
if ( nPenWidth % 2 == 0 )
m_pPainter->setRenderHint(QPainter::Qt4CompatiblePainting, true);