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

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

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  • H hobbyProgrammer
    21 Nov 2019, 11:36

    @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

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    JonB
    wrote on 21 Nov 2019, 11:41 last edited by
    #12

    @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++.

    1 Reply Last reply
    0
    • J J.Hilk
      21 Nov 2019, 11:38

      @hobbyProgrammer

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

      The definition is made automatically in generated code by moc

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      hobbyProgrammer
      wrote on 21 Nov 2019, 12:33 last edited by
      #13

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

      J J 2 Replies Last reply 21 Nov 2019, 12:36
      0
      • H hobbyProgrammer
        21 Nov 2019, 12:33

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

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        J.Hilk
        Moderators
        wrote on 21 Nov 2019, 12:36 last edited by J.Hilk
        #14

        @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();
                }
        }
        

        Be aware of the Qt Code of Conduct, when posting : https://forum.qt.io/topic/113070/qt-code-of-conduct


        Q: What's that?
        A: It's blue light.
        Q: What does it do?
        A: It turns blue.

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        • H hobbyProgrammer
          21 Nov 2019, 12:33

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

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          JonB
          wrote on 21 Nov 2019, 12:37 last edited by JonB
          #15

          @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().

          H 1 Reply Last reply 21 Nov 2019, 12:53
          0
          • J JonB
            21 Nov 2019, 12:37

            @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().

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            hobbyProgrammer
            wrote on 21 Nov 2019, 12:53 last edited by
            #16

            @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?

            J 1 Reply Last reply 21 Nov 2019, 13:04
            0
            • H hobbyProgrammer
              21 Nov 2019, 12:53

              @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?

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              JonB
              wrote on 21 Nov 2019, 13:04 last edited by
              #17

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

              H 1 Reply Last reply 21 Nov 2019, 13:09
              0
              • J JonB
                21 Nov 2019, 13:04

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

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                hobbyProgrammer
                wrote on 21 Nov 2019, 13:09 last edited by
                #18

                @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);
                
                J 1 Reply Last reply 21 Nov 2019, 13:13
                0
                • H hobbyProgrammer
                  21 Nov 2019, 13:09

                  @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);
                  
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                  jsulm
                  Lifetime Qt Champion
                  wrote on 21 Nov 2019, 13:13 last edited by
                  #19

                  @hobbyProgrammer Did you make sure

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

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

                  https://forum.qt.io/topic/113070/qt-code-of-conduct

                  H 1 Reply Last reply 21 Nov 2019, 13:18
                  1
                  • J jsulm
                    21 Nov 2019, 13:13

                    @hobbyProgrammer Did you make sure

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

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

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                    hobbyProgrammer
                    wrote on 21 Nov 2019, 13:18 last edited by
                    #20

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

                        LoginWidget *login = qobject_cast<LoginWidget*>(ui->stackedWidget->widget(0));
                    
                    J 1 Reply Last reply 21 Nov 2019, 13:19
                    0
                    • H hobbyProgrammer
                      21 Nov 2019, 13:18

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

                          LoginWidget *login = qobject_cast<LoginWidget*>(ui->stackedWidget->widget(0));
                      
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                      jsulm
                      Lifetime Qt Champion
                      wrote on 21 Nov 2019, 13:19 last edited by
                      #21

                      @hobbyProgrammer And connect succeeded?

                      https://forum.qt.io/topic/113070/qt-code-of-conduct

                      H 1 Reply Last reply 21 Nov 2019, 13:24
                      1
                      • J jsulm
                        21 Nov 2019, 13:19

                        @hobbyProgrammer And connect succeeded?

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                        hobbyProgrammer
                        wrote on 21 Nov 2019, 13:24 last edited by
                        #22

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

                        J 1 Reply Last reply 21 Nov 2019, 13:26
                        0
                        • H hobbyProgrammer
                          21 Nov 2019, 13:24

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

                          J Online
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                          jsulm
                          Lifetime Qt Champion
                          wrote on 21 Nov 2019, 13:26 last edited by
                          #23

                          @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);
                          

                          https://forum.qt.io/topic/113070/qt-code-of-conduct

                          H 1 Reply Last reply 21 Nov 2019, 13:27
                          2
                          • J jsulm
                            21 Nov 2019, 13:26

                            @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);
                            
                            H Offline
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                            hobbyProgrammer
                            wrote on 21 Nov 2019, 13:27 last edited by
                            #24

                            @jsulm yes thank you. That worked!

                            Thank you so much for you patience and help.

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                            • J Offline
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                              J.Hilk
                              Moderators
                              wrote on 21 Nov 2019, 13:27 last edited by
                              #25

                              Because this is getting out of hand:

                              https://github.com/DeiVadder/LoginExample


                              Be aware of the Qt Code of Conduct, when posting : https://forum.qt.io/topic/113070/qt-code-of-conduct


                              Q: What's that?
                              A: It's blue light.
                              Q: What does it do?
                              A: It turns blue.

                              1 Reply Last reply
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                              21 Nov 2019, 13:19

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