I'm trying to show a start-up image using QSplashScreen and I want to show the image for about 2 seconds.
int main(int argc, char *argv[]) {
QApplication a(argc, argv);
QPixmap pixmap(":/images/usm.png");
QSplashScreen splash(pixmap);
splash.show();
splash.showMessage("Loading Processes");
QTimer::singleShot(2000, &splash, SLOT(close()));
MainWindow w;
w.show();
splash.finish(&w);
return a.exec();
}
But this is not working. QSplashScreen appears for some milliseconds and then disappears. Tried to modify the time period but it seems like the QSplashScreen object is not connected to slot. What's the problem and how to avoid it?
The problem with your code is that the timer is not blocking the execution so the splash screen has already closed with the splash.finish(&w) call. What you need is a sleep. You could use a QWaitCondition like this:
int main(int argc, char *argv[])
{
QApplication a(argc, argv);
QSplashScreen splash(QPixmap(":/images/usm.png"));
splash.show();
splash.showMessage("Loading Processes");
// Wait for 2 seconds
QMutex dummyMutex;
dummyMutex.lock();
QWaitCondition waitCondition;
waitCondition.wait(&dummyMutex, 2000);
MainWindow w;
w.show();
splash.finish(&w);
return a.exec();
}
The disadvantage of this approach is that you are blocking the execution. If you do not want to block it then you can simply remove the splash.finish(&w) call:
int main(int argc, char *argv[]) {
QApplication a(argc, argv);
QPixmap pixmap(":/images/usm.png");
QSplashScreen splash(pixmap);
splash.show();
splash.showMessage("Loading Processes");
QTimer::singleShot(2000, &splash, SLOT(close()));
MainWindow w;
w.show();
return a.exec();
}
This code should work:
int main(int argc, char *argv[])
{
QApplication a(argc, argv);
QSplashScreen splash(QPixmap(":/images/usm.png"));
splash.showMessage("Loading Processes");
splash->show();
QMainWindow w;
QTimer::singleShot(2000, splash, SLOT(close()));
QTimer::singleShot(2500, &w, SLOT(show()));
return a.exec();
}
Related
Whenever the application window is minimized or maximized i want to link that signal with a function.
This is the code.
int main(int argc, char *argv[])
{
QApplication application(argc, argv);
Renderer w(model ); // This is QWidget
w.show();
QObject::connect(&w, &QWindow::windowStateChanged, [&](Qt::WindowState state) {
});
// how will i define the QObject::connect
return application.exec();
}
What would be the parameters for the QObject::connect function ?
You cannot use the connect function to connect to different slots based on the given value. You can however simply call the functions based on the value by checking the value in your lambda.
At least, you could if you had the signal. However, your connect suggests that w is - or inherits - a QWindow. You can obviously only connect to signals your class provides. As your Renderer is a QWidget, you have to check that class.
The documentation of QWidget tells us, that there is no windowStateChanged signal, but it states:
When the window state changes, the widget receives a changeEvent() of type QEvent::WindowStateChange.
So therefor we can create our own signal and connect to that. This can look similar to the following working example:
#ifndef RENDERER_H
#define RENDERER_H
#include <QWidget>
#include <QEvent>
class Renderer : public QWidget {
Q_OBJECT
signals:
void stateChanged(bool isMaximized);
protected:
void changeEvent(QEvent *e)
{
if(e->type() == QEvent::WindowStateChange) {
emit stateChanged(windowState() & ~Qt::WindowMaximized);
}
QWidget::changeEvent(e);
}
};
#endif // RENDERER_H
int main(int argc, char *argv[])
{
QApplication application(argc, argv);
Renderer w; // This is QWidget
w.show();
QObject::connect(&w, &Renderer::stateChanged, [&](bool maximized) {
qDebug() << "Maximized?" << maximized;
});
return application.exec();
}
I was able to solve by using QApplication::focusWindow()
int main(int argc, char *argv[])
{
QApplication application(argc, argv);
Renderer w; // This is QWidget
w.show();
QObject::connect(QApplication::focusWindow(), &Renderer::stateChanged, [&](bool maximized) {
qDebug() << "Maximized?" << maximized;
});
return application.exec();
}
Got an error running the following problem:
#include <QMessageBox>
#include <QApplication>
int main() {
QApplication app();
QMessageBox msgBox(QMessageBox::Critical,
QObject::tr("text1"),
QObject::tr("text2"),
QMessageBox::Ok);
msgBox.exec();
return 0;
}
The error is:
The program breaks at QMessageBox msgBox(...);
The call stack is:
EDIT: Even after i have added QApplication instance in XTerm window named qtcreator_process_stub i see the following:
QWidget: Must construct QApplication before a QWidget
The line
QApplication app();
doesn't create a QApplication object - it actually declares a function taking no arguments and returning a QApplication! This is sometimes known as the "most vexing parse".
To actually construct the application object, you need to provide the program's arguments:
QApplication app(argc, argv);
The full program is then
#include <QMessageBox>
#include <QApplication>
int main(int argc, char **argv) {
QApplication app(argc, argv);
QMessageBox msgBox(QMessageBox::Critical,
QObject::tr("text1"),
QObject::tr("text2"),
QMessageBox::Ok);
msgBox.exec();
return 0;
}
i want to remove the icon on the top left of a qt application. here is my main.cpp code
int main(int argc, char *argv[])
{
QCoreApplication::setAttribute(Qt::AA_EnableHighDpiScaling);
QGuiApplication app(argc, argv);
QQmlApplicationEngine engine;
engine.load(QUrl(QLatin1String("qrc:/main.qml")));
Receive receive;
QObject *topLevel = engine.rootObjects().value(0);
QQuickWindow *window = qobject_cast<QQuickWindow *>(topLevel);
// setWindowFlags(Qt::Dialog | Qt::CustomizeWindowHint | Qt::WindowTitleHint);
QObject::connect(window,SIGNAL(sendUrltoC(QUrl)),&receive,SLOT(getText(QUrl)));
return app.exec();
}
if i uncomment the commented part it gives me an error. what should i do?
Thank you for going through this post. I have searched forums,blogs and SO, but could not get what I actually need.
I am experimenting on how to display multiple mainwindows. I am using a embedded hardware board. I have successfully ported QT lib on to it.
I have written a small program.
mainwindow.cpp
MainWindow::MainWindow(QWidget *parent) :
QMainWindow(parent),
ui(new Ui::MainWindow)
{
ui->setupUi(this);
}
MainWindow::~MainWindow()
{
delete ui;
}
main.cpp
int main(int argc, char *argv[])
{
QApplication a(argc, argv);
MainWindow text_plane;
text_plane.setWindowFlags(Qt::Window | Qt::FramelessWindowHint);
text_plane.setStyleSheet("background-color: Black;");
text_plane.show();
a.exec();
return a.exec();
}
The above code displays one window only.
Even if I create a Mainwindow w1 after text_plane.show() like
int main(int argc, char *argv[])
{
QApplication a(argc, argv);
MainWindow text_plane,w1;
text_plane.setWindowFlags(Qt::Window | Qt::FramelessWindowHint);
text_plane.setStyleSheet("background-color: Black;");
text_plane.show();
w1.setWindowFlags(Qt::Window | Qt::FramelessWindowHint);
w1.setStyleSheet("background-color: Yellow;");
w1.show();
a.exec();
return a.exec();
}
Now only the w1 window is shown. What about the text_plane window ?? how to get that back.
Can anyone help me out here to make this understand.
Thank you
First of all, I wonder, what are you trying to achieve here:
a.exec();
return a.exec();
You need to call this method only once, you know.
Second of all, both of your windows are shown (you can see it in a taskbar), but, since you set Qt::FramelessWindowHint, one of them is drawn on top of the other. You can split them by using move (or something like it):
int main(int argc, char *argv[])
{
QApplication app(argc, argv);
MainWindow w, w1;
w.setWindowFlags(Qt::Window | Qt::FramelessWindowHint);
w.setStyleSheet("background-color: Black;");
w.show();
w1.setWindowFlags(Qt::Window | Qt::FramelessWindowHint);
w1.setStyleSheet("background-color: Yellow;");
w1.show();
w.move(0, 0);
w1.move(100, 100);
return app.exec();
}
How do I have the user first login before getting to the main window in my QT app? I'm new to QT and have looked through the documentation and have not found anything very helpful. Thank you.
I would make this in the following way. Let's assume, that my login dialog is a QDialog:
class Login : public QDialog
{
[..]
};
In my application, I create an instance of my Login dialog and if it is accepted, i.e. closed with OK button, I open my main window:
int main(int argc, char *argv[])
{
[..]
QMainWindow mw;
Login login;
if (login.exec() == QDialog::Accepted) {
mw.show();
}
[..]
}
int main(int argc, char *argv[])
{
QApplication a(argc, argv);
MainWindow w;
LoginWindow l; // create login-widget
connect(&l, SIGNAL(loggedIn), &w, SLOT(show()); // connect to mainwindow show()
l.show(); // show login-window instead of
// mainwindow at start
return a.exec();
}
You might want to :
1 - exec the loop only if the dialog is accepted, otherwise you application continues to run
2 - create the mainwindow only if accepted, cause it might be a heavy interface, which takes time to initialize
int main(int argc, char *argv[])
{
QApplication a(argc, argv);
Dialog_Login dialog;
if(dialog.exec() == QDialog::Accepted){
MainWindow w;
w.setUser(dialog.GetUser());
w.show();
return a.exec();
}
else return 0;
}
You need to create the QApplication instance in parrallel with the login dialog.