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

[JonB]

@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))

[hobbyProgrammer]

@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

[J.Hilk]

@hobbyProgrammer

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

The definition is made automatically in generated code by moc

[JonB]

@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++.

[hobbyProgrammer]

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

[J.Hilk]

@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();
        }
}

[JonB]

@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().

[hobbyProgrammer]

@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?

[JonB]

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

[hobbyProgrammer]

@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);

[jsulm]

@hobbyProgrammer Did you make sure

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

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

[hobbyProgrammer]

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

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

[jsulm]

@hobbyProgrammer And connect succeeded?

[hobbyProgrammer]

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

[jsulm]

@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]

@jsulm yes thank you. That worked!

Thank you so much for you patience and help.

[J.Hilk]

Because this is getting out of hand:

https://github.com/DeiVadder/LoginExample