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

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  • H hobbyProgrammer

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

    JonBJ Offline
    JonBJ Offline
    JonB
    wrote on 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
    0
    • JonBJ 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().

      H Offline
      H Offline
      hobbyProgrammer
      wrote on 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?

      JonBJ 1 Reply Last reply
      0
      • H 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?

        JonBJ Offline
        JonBJ Offline
        JonB
        wrote on last edited by
        #17

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

        H 1 Reply Last reply
        0
        • JonBJ JonB

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

          H Offline
          H Offline
          hobbyProgrammer
          wrote on 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);
          
          jsulmJ 1 Reply Last reply
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          • H 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);
            
            jsulmJ Offline
            jsulmJ Offline
            jsulm
            Lifetime Qt Champion
            wrote on 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
            1
            • jsulmJ jsulm

              @hobbyProgrammer Did you make sure

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

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

              H Offline
              H Offline
              hobbyProgrammer
              wrote on last edited by
              #20

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

                  LoginWidget *login = qobject_cast<LoginWidget*>(ui->stackedWidget->widget(0));
              
              jsulmJ 1 Reply Last reply
              0
              • H hobbyProgrammer

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

                    LoginWidget *login = qobject_cast<LoginWidget*>(ui->stackedWidget->widget(0));
                
                jsulmJ Offline
                jsulmJ Offline
                jsulm
                Lifetime Qt Champion
                wrote on last edited by
                #21

                @hobbyProgrammer And connect succeeded?

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

                H 1 Reply Last reply
                1
                • jsulmJ jsulm

                  @hobbyProgrammer And connect succeeded?

                  H Offline
                  H Offline
                  hobbyProgrammer
                  wrote on last edited by
                  #22

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

                  jsulmJ 1 Reply Last reply
                  0
                  • H hobbyProgrammer

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

                    jsulmJ Offline
                    jsulmJ Offline
                    jsulm
                    Lifetime Qt Champion
                    wrote on 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
                    2
                    • jsulmJ 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);
                      
                      H Offline
                      H Offline
                      hobbyProgrammer
                      wrote on last edited by
                      #24

                      @jsulm yes thank you. That worked!

                      Thank you so much for you patience and help.

                      1 Reply Last reply
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                      • J.HilkJ Offline
                        J.HilkJ Offline
                        J.Hilk
                        Moderators
                        wrote on 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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