QT Quick no events ("onClicked" or "onCurrentIndexChanged") work - what I am doing wrong?

[sm-a]

I am new to QT Quick and have a problem what I can't solve with Google search. I hope I get here help.

My problem: I have created a QT Quick App from the template "QT Quick Application Stack". But no "onClicked" event or "onCurrentIndexChanged" event will be executed. What I am doing wrong?

Test.qml

import QtQuick 2.9
import QtQuick.Dialogs 1.2

TestForm {
    button.onClicked: {
        dialog.open()
    }
    countryComboBox.onCurrentIndexChanged: {
        dialog.open()
    }

    MessageDialog {
                id: dialog
                title: qsTr("this is a test")
                text: qsTr("test ok")
                
                function show(caption) {
                    dialog.open();
               }
    }
}

TestForm.ui.qml

import QtQuick 2.9
import QtQuick.Controls 2.2

Page {
    width: 600
    height: 400
    property alias button: button
    property alias countryComboBox: countryComboBox

    title: qsTr("Country select")

    ComboBox {
        currentIndex: 0
        id: countryComboBox
        x: 16
        y: 12

        model: ["Banana", "APPLE"]
    }

    Button {
        id: button
        x: 132
        y: 102
        text: qsTr("Button")
    }
}

[gfxx]

on designer inside ui.qml file ... design mode ... hover the button .... right click on mouse button ... choose "go to implementation".

without these passages your code not work. After these you find in your qml file

            button.onClicked: {
          console.log("log completed")
}

regards

I just try to write the code without use the mouse trick .... and it not work.

[sm-a]

Thank you @gfxx,

I tried your method, but it still doesn't work.

I have the Stack template in use and I have added a "QT Quick UI file" with the name "TestForm". When I tried to implement this with "go to implementation" in the "TestForm" it dosen't work.

Do you have any idea why this might be?

Greetings,
Sven

[gfxx]

@sm-a Have you try to do sometings like these?

in your main.qml

 SwipeWiew:{
       /* some layout code */

         YourFirstPageForm:{
                     /* some layout code*/
                     function yourfunction(){
                             console.log("OK these work!!")
                   }
yourqmlwidget.onClick: yourfunction()
                }
         YourSecondPageForm:{
              }
   }

/* some code for navigate*/

These still not work? I attach some image (first, second, third step png.) So you can see my sample project ....

regards

[sm-a]

@gfxx very thank you this helps. But how can I define it to use "Test.qml" (from "TestForm.ui.qml") instead of "main.qml" for the definition?

[gfxx]

I try to attach my sample project ... (work to my home for techer me about qml and quick world that not know too) I am attaching my example project (which I also use to learn at home qml and quick that I do not know at all well ..). This for me works very well for what you ask ... I have problems understanding how to use tableView and qmlregistertype .... but it's another story.

These because my english is poor so with the code explain me better.

I not have the required perission .... so i use git these is my folder project

Attention: in disegn mode, navigator panel control the circle with arrow state ... if not red ... the action is not performed

I hope these help

regards

[shaan7]

Hi,

I followed the same steps as you and it works for me. My project files are like this-

/tmp/test$ tree
.
├── main.cpp
├── main.qml
├── qml.qrc
├── qtquickcontrols2.conf
├── TestForm.ui.qml
├── test.pro
├── test.pro.user
└── Test.qml

main.qml

import QtQuick 2.12
import QtQuick.Controls 2.5

ApplicationWindow {
    id: window
    visible: true
    width: 640
    height: 480
    title: qsTr("Stack")

    StackView {
        id: stackView
        initialItem: "Test.qml"
        anchors.fill: parent
    }
}

[gfxx]

@ sm-a is not necessary the name "main" in main.qml .... you must declare it on qrc file and insert in on main.cpp. main.qml is the default name on qt designer system.

The default name is not necessary at all.... testuiform can be MyIceScream.ui.qml for example .... as said @shaan7

If you have solved your issue use "solved button"

regards

[sm-a]

Many thanks to everyone who helped me (@shaan7 and @gfxx) it now works as it should.

[gfxx]

@sm-a with pleasure ...