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QStackedWidget: How to check if the next widget should be shown



  • I've made an application with a stacked widget. In the MainWindow the widgets are set. The first widget is a login page, after a succesfull login I'd like to show the second page. How can I check inside the MainWindow if the login was succesfull? (Both widgets are classes with ui's) I also promoted the widgets of the stackedWidget to these classes.

    this is my code:

    MainWindow.cpp

    #include "mainwindow.h"
    #include "loginwidget.h"
    #include "ui_mainwindow.h"
    
    MainWindow::MainWindow(QWidget *parent)
        : QMainWindow(parent)
        , ui(new Ui::MainWindow)
    {
        ui->setupUi(this);
        ui->stackedWidget->setCurrentIndex(0);
        
    }
    
    MainWindow::~MainWindow()
    {
        delete ui;
    }
    

    loginWidget.cpp

    #include "loginwidget.h"
    #include "ui_loginwidget.h"
    
    LoginWidget::LoginWidget(QWidget *parent) :
        QWidget(parent),
        ui(new Ui::LoginWidget)
    {
        ui->setupUi(this);
    }
    
    LoginWidget::~LoginWidget()
    {
        delete ui;
    }
    
    bool LoginWidget::loginSuccesfull()
    {
        QString username = ui->lineEdit_2->text();
        QString password = ui->lineEdit->text();
        
        if(username == "Test" && password == "Test123")
        {
            return true;
        }
        else
        {
            return false;
        }
    }
    

    AfterLoginScreen (currently a non-functional widget)

    #include "afterloginscreen.h"
    #include "ui_afterloginscreen.h"
    
    AfterLoginScreen::AfterLoginScreen(QWidget *parent) :
        QWidget(parent),
        ui(new Ui::AfterLoginScreen)
    {
        ui->setupUi(this);
    }
    
    AfterLoginScreen::~AfterLoginScreen()
    {
        delete ui;
    }
    

  • Lifetime Qt Champion

    @hobbyProgrammer said in QStackedWidget: How to check if the next widget should be shown:

    but I don't think so

    Then it can't work.
    Use new Qt5 connect syntax to be sure signal/slot are really connected:

    connect(login, &LoginWidget::loginSuccesfull, this, &MainWindow::showApp);
    


  • @hobbyProgrammer said in QStackedWidget: How to check if the next widget should be shown:

    How can I check inside the MainWindow if the login was succesfull? (

    Call your LoginWidget::loginSuccesfull()!

    If you mean, how do you get a reference to the login widget to call loginWidget->loginSuccesfull(), either keep a reference to the loginWidget at the time you create it/put it into the QStackedWidget, or use something like

    loginWidget = qobject_cast<LoginWidget*>(ui->stackedWidget.widget(1))
    


  • @JonB Oh I'm sorry, it should happen on a pushbutton (when it's pressed, it checks if the login was good)

    So like this:

    bool LoginWidget::loginSuccesfull()
    {
        QString username = ui->lineEdit_2->text();
        QString password = ui->lineEdit->text();
    
        if(username == "Test" && password == "Test123")
        {
            return true;
        }
        else
        {
            return false;
        }
    }
    
    void LoginWidget::on_pushButton_clicked()
    {
        loginSuccesfull();
    }
    

    but how do I check the value of loginSuccesfull() in the mainwindow?



  • @hobbyProgrammer
    There is no point invoking a function from on_pushButton_clicked() which just returns true or false without acting on the return result.

    There are many ways to achieve what you want. I'll just say that it would be easier if you chose to make the login page a modal dialog instead of one of the stacked widgets. That's a different interface, I don't know how you feel about that.

    If you want to keep the login in stacked widget as now: Probably, stop loginSuccessfull() from being a function returning a boolean. Instead, in the false case have it show the user an error message. In the true case, one of:

    • Have it switch back to some other window by calling QStackedWidget::setCurrentIndex(0) (it will have to have access to ui->stackedWidget).
    • Have it raise a signal to indicate successful logon, and the main window will have a slot for that so that the ui->stackedWidget->setCurrentIndex(0) can be done in the main window.

    Ah, I see at present there is only one stacked widget, the login one, so the first alternative may not make sense. If you want your main window code to know about the login result, use the signal/slot mechanism.



  • @JonB alright, so I should include the mainWindow.ui file aswell instead of only the loginWidget.ui?



  • @hobbyProgrammer
    No, for your case I would suggest LoginWidget::loginSuccesfull() on success should raise a signal, and MainWindow should place a slot on that signal to receive it.

    I'm sorry but I don't have time to write this all out for you, maybe someone else will. Have you read https://doc.qt.io/qt-5/signalsandslots.html, which is the core of how Qt uses signals & slots?



  • @JonB Thanks. That's very helpful!



  • @JonB so should it be something like this?

        connect(ui->stackedWidget->indexOf(0), SIGNAL(loginSuccesfull()), this, SLOT(showApp()));
    

    I might be close, but it's not working yet. Do you see what I am doing wrong?

    I also tried this:

        Login *login = ui->stackedWidget->indexOf(0);
        
        connect(login, SIGNAL(loginSuccesfull()), this, SLOT(showApp()));
    


  • @hobbyProgrammer
    Whatever else might be wrong, QStackedWidget::indexOf() is totally the wrong way round. As I wrote earlier, you will need to use QStackedWidget::widget(), and if you need a reference to it being a LoginWidget to access members of that class you will want something like:

    loginWidget = qobject_cast<LoginWidget*>(ui->stackedWidget.widget(0))
    


  • @JonB thank you.

        LoginWidget *login = qobject_cast<LoginWidget*>(ui->stackedWidget->widget(0));
    
        connect(login, SIGNAL(loginSuccesfull()), this, SLOT(showApp()));
    

    I currently get these errors (as loginSuccesfull is a public bool method):

    QObject::connect: No such signal LoginWidget::loginSuccesfull() in ..\stackedLogin\mainwindow.cpp:14
    QObject::connect:  (sender name:   'page')
    QObject::connect:  (receiver name: 'MainWindow')
    

    whenever I change loginSuccesfull() to a signal instead of public bool I get these errors:

    LNK2005:"public: bool _thiscall LoginWidget::loginSuccesfull(void)......" and LNK1169: one or more multiply defined symbols found


  • Moderators

    @hobbyProgrammer

    A Signal must not have a definition, only a declaration.

    The definition is made automatically in generated code by moc



  • @hobbyProgrammer
    In some shape or form you have not changed over the code to how signals/slots are declared and work in Qt. There are lots of example to read. You will need correct Qt declarations using signals & slots in your header files, and loginSuccesfull() won't be some bool function, it will emit the signal.

    I will leave others to help you with this, I don't even do Qt in C++.



  • @J-Hilk alright, but how would I check if the login is correct if there's no function definition?


  • Moderators

    @hobbyProgrammer

    the idea would be, that you emit the signal, as soon as your function verifies, that the login was successful !

    void LoginWidget::on_pushButton_clicked()
    {
           QString username = ui->lineEdit_2->text();
            QString password = ui->lineEdit->text();
        
            if(username == "Test" && password == "Test123")
            {
                 emit loginSuccessful();
            }
    }
    


  • @hobbyProgrammer
    Subject to @J-Hilk suggesting an alternative, as I wrote earlier retain your loginSuccesfull() as the slot for the pushbutton click. Have it emit a signal you define on successful validation of the widgets, the signal is a separate thing from your function which you must define as per the Qt/C++ rules.

    EDIT OK, @J-Hilk's code is the same thing, it's just that he has defined the pushbutton slot as on_pushButton_clicked() and the signal as logInSuccessful, so there is no longer any (non-signal) function named loginSuccesfull().



  • @JonB alright, so using that code and these lines:

        LoginWidget *login = qobject_cast<LoginWidget*>(ui->stackedWidget->widget(0));
        connect(login, SIGNAL(loginSuccesfull()), this, SLOT(showApp()));
    

    should result in showApp to happen anytime the login was succesfull?



  • @hobbyProgrammer
    I'm hoping so! Did you try it?



  • @JonB yes and it didn't work. I tried debugging but I can't seem to find what's going wrong. It hits the code where the emit loginSuccessfull() is set, but it doesn't go to the SLOT where it's connected to and it ends in :

    while (!d->exit.loadAcquire())
            processEvents(flags | WaitForMoreEvents | EventLoopExec);
    

  • Lifetime Qt Champion

    @hobbyProgrammer Did you make sure

    connect(login, SIGNAL(loginSuccesfull()), this, SLOT(showApp()));
    

    succeeded? And is this "login" the one you're actually showing?



  • @jsulm
    it should be since I'm doing this:

        LoginWidget *login = qobject_cast<LoginWidget*>(ui->stackedWidget->widget(0));
    

  • Lifetime Qt Champion

    @hobbyProgrammer And connect succeeded?



  • @jsulm I'm not sure, but I don't think so.


  • Lifetime Qt Champion

    @hobbyProgrammer said in QStackedWidget: How to check if the next widget should be shown:

    but I don't think so

    Then it can't work.
    Use new Qt5 connect syntax to be sure signal/slot are really connected:

    connect(login, &LoginWidget::loginSuccesfull, this, &MainWindow::showApp);
    


  • @jsulm yes thank you. That worked!

    Thank you so much for you patience and help.


  • Moderators

    Because this is getting out of hand:

    https://github.com/DeiVadder/LoginExample