Skip to content
  • Categories
  • Recent
  • Tags
  • Popular
  • Users
  • Groups
  • Search
  • Get Qt Extensions
  • Unsolved
Collapse
Brand Logo
  1. Home
  2. Qt Development
  3. General and Desktop
  4. QStackedWidget: How to check if the next widget should be shown
Forum Updated to NodeBB v4.3 + New Features

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

Scheduled Pinned Locked Moved Solved General and Desktop
qstackedwidgetqstackedwidgets
25 Posts 4 Posters 3.7k Views 1 Watching
  • Oldest to Newest
  • Newest to Oldest
  • Most Votes
Reply
  • Reply as topic
Log in to reply
This topic has been deleted. Only users with topic management privileges can see it.
  • J JonB
    21 Nov 2019, 10:31

    @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?

    H Offline
    H Offline
    hobbyProgrammer
    wrote on 21 Nov 2019, 10:32 last edited by
    #7

    @JonB Thanks. That's very helpful!

    1 Reply Last reply
    0
    • J JonB
      21 Nov 2019, 10:31

      @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?

      H Offline
      H Offline
      hobbyProgrammer
      wrote on 21 Nov 2019, 10:44 last edited by hobbyProgrammer
      #8

      @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()));
      
      J 1 Reply Last reply 21 Nov 2019, 11:06
      0
      • H hobbyProgrammer
        21 Nov 2019, 10:44

        @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()));
        
        J Offline
        J Offline
        JonB
        wrote on 21 Nov 2019, 11:06 last edited by JonB
        #9

        @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))
        
        H 1 Reply Last reply 21 Nov 2019, 11:36
        1
        • J JonB
          21 Nov 2019, 11:06

          @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))
          
          H Offline
          H Offline
          hobbyProgrammer
          wrote on 21 Nov 2019, 11:36 last edited by
          #10

          @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 J 2 Replies Last reply 21 Nov 2019, 11:38
          0
          • 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

            J Offline
            J Offline
            J.Hilk
            Moderators
            wrote on 21 Nov 2019, 11:38 last edited by
            #11

            @hobbyProgrammer

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

            The definition is made automatically in generated code by moc


            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.

            H 1 Reply Last reply 21 Nov 2019, 12:33
            3
            • 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

              J Offline
              J Offline
              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

                H Offline
                H Offline
                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?

                  J Offline
                  J Offline
                  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.

                  1 Reply Last reply
                  3
                  • 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?

                    J Offline
                    J Offline
                    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().

                      H Offline
                      H Offline
                      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?

                        J Offline
                        J Offline
                        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?

                          H Offline
                          H Offline
                          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);
                            
                            J Offline
                            J Offline
                            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?

                              H Offline
                              H Offline
                              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));
                                
                                J Offline
                                J Offline
                                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?

                                  H Offline
                                  H Offline
                                  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 Offline
                                    J Offline
                                    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
                                      H Offline
                                      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.

                                      1 Reply Last reply
                                      0
                                      • J Offline
                                        J Offline
                                        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
                                        4

                                        16/25

                                        21 Nov 2019, 12:53

                                        • Login

                                        • Login or register to search.
                                        16 out of 25
                                        • First post
                                          16/25
                                          Last post
                                        0
                                        • Categories
                                        • Recent
                                        • Tags
                                        • Popular
                                        • Users
                                        • Groups
                                        • Search
                                        • Get Qt Extensions
                                        • Unsolved