"Don't show this meesage again" checkbox in QMessagebox

jsulm

@Prathamesh said in "Don't show this meesage again" checkbox in QMessagebox:

Is it correct?

Apparently it is not.
You are not doing what I suggested, which is actually easy...

Prathamesh

@jsulm Ok,
state = check_box.checkState()
if state == Qt.Checked:
pass

Now is it ok or am I still missing something? because it is appearing again..

jsulm

@Prathamesh said in "Don't show this meesage again" checkbox in QMessagebox:

if state == Qt.Checked:
pass

What do you want to achieve with this useless code?
Don't you think you should also check the stored state of the check box BEFORE showing ther dialog? As I suggested?

JonB

@Prathamesh
And in addition to @jsulm's comment: if you want to know the user previously checked the box so that you don't show the dialog again, you are going to have save somewhere in persistent state that the user previously checked that box. You can't get that from the dialog previously, because you create a new dialog each time.

Prathamesh

@JonB Checked all the methods given in documentation, tried to store checkbox status in a variable. But same thing is happening.

JonB

@Prathamesh It's your code then.

J.Hilk

Here's one way to do something like this:

int main(int argc, char *argv[])
{

    QCoreApplication::setAttribute(Qt::AA_EnableHighDpiScaling);
    QCoreApplication::setOrganizationName("MyCompany");
    QCoreApplication::setOrganizationDomain("MyApplication");
    QApplication app(argc, argv);


    QSettings s;

    QMessageBox msgB(QMessageBox::Critical,"One TimeBox", "This MessageBox, when confirmed will never be shown again");

    msgB.setStandardButtons(QMessageBox::Ok | QMessageBox::Cancel);

    QTimer t;
    t.setSingleShot(true);

    if(!s.value("OneTimeOK",false).toBool()){
        
        QObject::connect(&t, &QTimer::timeout,&msgB, &QMessageBox::exec);

        QObject::connect(&msgB, &QMessageBox::finished, [&s,&t, &app](int result){
            qDebug() << "Finished" << result;
            if(result == QMessageBox::Ok){
                s.setValue("OneTimeOK", true);
                app.quit();
            } else {
                t.start(1000);
            }

        });
        msgB.exec();
    } else {
        qDebug() << "Nothing to show";
        return -1;
    }

    return app.exec();
}

Prathamesh

@JonB

def __gui_initial_state(self):
    """
    This method initializes the gui state.
    """
    self.initial_state.emit()
    self.__update_recent_configurations()
    version_str, version_tuple = global_vars.check_latest_version()
             
    if global_vars.LOCAL_VERSION_TUPLE >= version_tuple:
        msg = QtWidgets.QMessageBox()
        msg.setWindowTitle("What's New")
        msg.setWindowFlags(QtCore.Qt.WindowStaysOnTopHint)
        msg.setWindowIcon(QtGui.QIcon(os.path.abspath(os.path.join(os.path.dirname(
            os.path.realpath(__file__)), 'gui_images', 'logo_icon.png'))))
        msg.setText(ui_helper.WHAT_IS_NEW)
        check_box = QtWidgets.QCheckBox("Don't show this again")
        msg.setCheckBox(check_box)
        msg.exec_()
        check_box_status = check_box.isChecked()
        if check_box_status:
            print("abcd")

This method executes everytime when a gui in launched. My task is to check whether new version of framework is launched or not, and if launched user will download it. After downloading my code will check if it is downloaded and then a pop-up containing all the information about new version will come up.

So,this pop-up will appear first time when user downloads new version, and then in this pop-up i have to put a checkbox. So I have done that,and using ischecked() method its status is getting stored in a variable and as i have written print("abcd") if status is true. So it is working fine.

But the problem is, everytime a gui is launched this pop-up will come up,so even after storing True status from previos run ,it is appearing. I am not understanding how to handle this situation.

jsulm

@Prathamesh said in "Don't show this meesage again" checkbox in QMessagebox:

But the problem is, everytime a gui is launched this pop-up will come up,so even after storing True status from previos run

Of course, you are still not doing what was suggested!

def __gui_initial_state(self):
    """
    This method initializes the gui state.
    """
    self.initial_state.emit()
    self.__update_recent_configurations()
    version_str, version_tuple = global_vars.check_latest_version()
             
    if global_vars.LOCAL_VERSION_TUPLE >= version_tuple and not self.__check_box_status:
        msg = QtWidgets.QMessageBox()
        msg.setWindowTitle("What's New")
        msg.setWindowFlags(QtCore.Qt.WindowStaysOnTopHint)
        msg.setWindowIcon(QtGui.QIcon(os.path.abspath(os.path.join(os.path.dirname(
            os.path.realpath(__file__)), 'gui_images', 'logo_icon.png'))))
        msg.setText(ui_helper.WHAT_IS_NEW)
        check_box = QtWidgets.QCheckBox("Don't show this again")
        msg.setCheckBox(check_box)
        msg.exec_()
        self.__check_box_status = check_box.isChecked()
        if self.__check_box_status:
            print("abcd")

Prathamesh

@jsulm I did half part of storing status.But I am not getting how to check it while launching a gui next time? That is where I am stuck. Any suggesstion please?

jsulm

@Prathamesh said in "Don't show this meesage again" checkbox in QMessagebox:

how to check it while launching a gui next time?

What exactly do you mean by that? Do you mean you close the app and then start it again and you want to check this check status? If so, then use QSettings to store such information permanently.

Prathamesh

@jsulm yeah this is what I mean exactly. Once I tick a checkbox and close an app then pop-up should not appear until new version of framework is launched.

jsulm

@Prathamesh https://doc.qt.io/qt-5/qsettings.html

Pl45m4

@Prathamesh

Then you need to store something like QSettings.
So that your app will know what your settings from your last run are.

Prathamesh

@jsulm Thanks for this. I have tried this just now. I have written something like this

--> msg = QtWidgets.QMessageBox()
msg.setWindowTitle("What's New")
msg.setWindowFlags(QtCore.Qt.WindowStaysOnTopHint)
msg.setText(ui_helper.WHAT_IS_NEW)
check_box = QtWidgets.QCheckBox("Don't show this again")
msg.setCheckBox(check_box)
msg.exec_()
check_box_status = check_box.isChecked()
settings = QtCore.QSettings()
settings.setValue(check_box_status, True)

I know this is not the correct way of doing and that's why I am getting an error --> setValue has unexpected type 'bool'. But then how do I pass the status of checkbox into this function?

SGaist

Hi,

Please read the function documentation. There are examples shown there on how it works.

QSettings works on the key/value paradigm.

jsulm

@Prathamesh said in "Don't show this meesage again" checkbox in QMessagebox:

settings.setValue(check_box_status, True)

Please read documentation to know how to do it properly, there are even code examples...

Prathamesh

@jsulm So I have read the documentation and from that i got to know that QSettings will store the value in registry editor in our machine. So I have written a code like below-->

msg = QtWidgets.QMessageBox()
msg.setWindowTitle("What's New")
msg.setWindowFlags(QtCore.Qt.WindowStaysOnTopHint)
msg.setText(ui_helper.WHAT_IS_NEW)
check_box = QtWidgets.QCheckBox("Don't show this again")
msg.setCheckBox(check_box)
msg.exec_()
check_box_status = check_box.isChecked()
settings = QtCore.QSettings('ConTest', 'Check-Box status')
settings.setValue("check_box_status", check_box_status)

Now I can see the status of checkbox as true or false in the registry. I need help here. Whatever I have written in my code is correct or not? Because the pop-up is still appearing ,that means I am not making use of registry data. How can I do that in my code?

jsulm

@Prathamesh said in "Don't show this meesage again" checkbox in QMessagebox:

Because the pop-up is still appearing

Of course it does. You do not check what you stored in settings to decide whether you should show the dialog or not...

JonB

@Prathamesh
So your code now stores the desired value in the registry. What do you now do about reading that value back, and then acting on it for (not showing) the dialog?