How to show/hide qml widget from c++ like BusyIndicator

[jigarp]

Hello,
I wanted to change visible status of Busyindicator widget where I'm trying to change it from c++. Where BackEnd is my c++ class.

    BackEnd {
        id: backend
    }
    BusyIndicator {
        id: busyIndicator
        x: 1011
        y: 332
        running: backend.visibleState
        onRunningChanged: backend.visibleState = running
    }

Here is my BackEnd.cpp

void BackEnd::setUserState(bool stateSet)
{
    state = !stateSet;
    emit userStateChanged();
}
bool BackEnd::visibleState()
{
    return state;
}
void BackEnd::timerTimeOut()
{
    setUserState(state);
}

Here is BackEnd.h

class BackEnd : public QObject
{
    Q_OBJECT
    Q_PROPERTY(bool visibleState READ visibleState WRITE setUserState NOTIFY userStateChanged)

public:
    explicit BackEnd(QObject *parent = nullptr);

    bool visibleState();
    void setUserState(bool stateSet);

signals:
    void userStateChanged();

private slots:
    void timerTimeOut();

private:
    bool state = false;
    QTimer * timer1;
};

Here I can see busy indicator show/hide whenever timeout()
But shows warning

QML BusyIndicator: Binding loop detected for property "running"

What is the better way to show or hide QML widget from C++

Thanks

[J.Hilk]

hi @jigarp

this is untested, but you could try this

BackEnd {
        id: backend
    }
    BusyIndicator {
        id: busyIndicator
        x: 1011
        y: 332
        visible: backend.visibleState
        running: visible
    }

should hide the busy indicator when it's not running.

[jigarp]

@J.Hilk

This stops error but not updating widget on screen.

[J.Hilk]

@jigarp what's the intervall of the timer ?

also do you see consol logs that tell you the indicator should be changed ?

BackEnd {
        id: backend
        onVisibleStateChanged: console.log("Indicator should be visible", visibleState)
    }

[jigarp]

@J.Hilk

My Timer ticks at every 1 second.

onVisibleStateChanged: console.log("Indicator should be visible",visibleState)

prints Indicator should be visible false at every 1 second

[J.Hilk]

@jigarp
very strange, your state does not change to ture. !?

try the following

void BackEnd::setUserState(bool stateSet)
{
    if( stateSet != state) {
           state = stateSet;
           emit userStateChanged();
     }
}

void BackEnd::timerTimeOut()
{
    setUserState(!visibleState());
}

[GrecKo]

@jigarp said in How to show/hide qml widget from c++ like BusyIndicator:

running: backend.visibleState
onRunningChanged: backend.visibleState = running

the running: backend.visibleState line should be enough, what's the point with : onRunningChanged: backend.visibleState = running ?

[Yaswanth]

This should work, check the following,

In BackEnd.h:

#include <QObject>
#include <QTimer>
#include <QDebug>

class BackEnd : public QObject
{
    Q_OBJECT
    Q_PROPERTY(bool visibleState READ visibleState WRITE setUserState NOTIFY userStateChanged)

public:
    BackEnd();
    bool visibleState();
    void setUserState(bool stateSet);

signals:
    void userStateChanged();

private slots:
    void timerTimeOut();

private:
    bool state = false;
    QTimer * timer1;
};

In BackEnd.cpp, I have given interval of 1 sec (1000 msec),

#include "BackEnd.h"

BackEnd::BackEnd()
{
    timer1 = new QTimer();
    connect(timer1,SIGNAL(timeout()),this,SLOT(timerTimeOut()));
    timer1->setInterval(1000);
    timer1->start();
}

void BackEnd::setUserState(bool stateSet)
{
    state = !stateSet;
    qDebug()<<"Current state is "<<state;
    emit userStateChanged();
}

bool BackEnd::visibleState()
{
    return state;
}

void BackEnd::timerTimeOut()
{
    qDebug()<<"Timer triggered";
    setUserState(state);
}

In Qml:
    BackEnd {
       id: backend
    }

    BusyIndicator {
       id: busyIndicator
       width: 100
       height: 100
       x: 100
       y: 100
       running: backend.visibleState
    }