Now, I have 1 application, but I don't want to open application twice, so I using QShareMemory
to detect application when open twice. And my question is: how I show current application in screen when user open application the second ?
int main(int argc, char *argv[]) {
Application a(argc, argv);
/*Make sure only one instance of application can run on host system at a time*/
QSharedMemory sharedMemory;
sharedMemory.setKey ("Application");
if (!sharedMemory.create(1))
{
qDebug() << "123123Exit already a process running";
return 0;
}
/**/
return a.exec();
}
Thanks.
Here's another approach in pure Qt way:
Use QLocalServer
and QLocalSocket
to check the existence of application and then use signal-slot mechanism to notify the existing one.
#include "widget.h"
#include <QApplication>
#include <QObject>
#include <QLocalSocket>
#include <QLocalServer>
int main(int argc, char *argv[])
{
QApplication a(argc, argv);
const QString appKey = "applicationKey";
QLocalSocket *socket = new QLocalSocket();
socket->connectToServer(appKey);
if (socket->isOpen()) {
socket->close();
socket->deleteLater();
return 0;
}
socket->deleteLater();
Widget w;
QLocalServer server;
QObject::connect(&server,
&QLocalServer::newConnection,
[&w] () {
/*Set the window on the top level.*/
w.setWindowFlags(w.windowFlags() |
Qt::WindowStaysOnTopHint);
w.showNormal();
w.setWindowFlags(w.windowFlags() &
~Qt::WindowStaysOnTopHint
);
w.showNormal();
w.activateWindow();
});
server.listen(appKey);
w.show();
return a.exec();
}
But if you're using Qt 5.3 on Windows, there's a bug for QWidget::setWindowFlags
and Qt::WindowStaysOnTopHint
, see https://bugreports.qt.io/browse/QTBUG-30359 .
Just use QSingleApplication
class instead of QApplication
: https://github.com/qtproject/qt-solutions/tree/master/qtsingleapplication
int main(int argc, char **argv)
{
QtSingleApplication app(argc, argv);
if (app.isRunning())
return 0;
MyMainWidget mmw;
app.setActivationWindow(&mmw);
mmw.show();
return app.exec();
}
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