触发时无法对函数进行 link QAction (Qt 5)

Question

我正在尝试将上下文菜单添加到系统托盘（可通过单击系统托盘图标激活）我已成功添加菜单和带有文本的操作 "Exit"，但我不知道如何操作到 link 动作 "triggered" 功能到另一个功能/更改触发功能或其他任何有效的功能。我只是想在单击操作时激活某些特定行为。当我单击此操作按钮时，它不会执行任何操作。我尝试 link 使用此构造函数将其传递给成员函数：QAction *QMenu::addAction(const QIcon &icon, const QString &text, const QObject *receiver, PointerToMemberFunction method, const QKeySequence &shortcut = ...) 这是我的代码中最重要的部分：

     MainWindow::MainWindow(QWidget *parent) :
        QMainWindow(parent),
        ui(new Ui::MainWindow)
    {
        ui->setupUi(this);
        mSystemTrayIcon = new QSystemTrayIcon(this);
        mSystemTrayIcon->setIcon(QIcon(":/iris_logo.png"));
        mSystemTrayIcon->setVisible(true);

        systemTrayMenu = new QMenu("Context menu");
        systemTrayMenu->setToolTipsVisible(true);
 // I get the error: no matching member function for call to 'addAction'
        systemTrayMenu->addAction("Open", this, on_actionQuit_triggered())); 

// I dont get an error, however this only creates a menu button, not its corresponding function that must be called.
        systemTrayMenu->addAction("Exit"); 
        mSystemTrayIcon->setContextMenu(systemTrayMenu);
    }

Answer 1

addAction returns指向QAction对象的指针，得到这个指针并使用connect在triggered信号和[=15之间建立连接=]插槽：

QAction* openAction = systemTrayMenu->addAction("Open"); 
connect (openAction, SIGNAL(triggered()) , this, SLOT(on_actionQuit_triggered()));

触发时无法对函数进行 link QAction (Qt 5)

Unable to link QAction to a function when triggered (Qt 5)

c++

qt5

qaction