只需按一下即可将QToolButton移到不同的布局？

Question

I have here two different layouts, and one QToolButton. My goal is to transfer that button between the two layouts when I click it. I figured this code would work,

snippet:

void DominionLinux::on_toolButton_clicked(string state)
{
    if (state=="Disabled"){
        ui->verticalLayout_Enabled->addWidget(ui->toolButton);
        state = "Enabled";
    }
    else if (state=="Enabled"){
        ui->verticalLayout_Disabled->addWidget(ui->toolButton);
        state = "Disabled";
    }
}

By default, state == "Disabled". When I test the UI in QTCreator, the first time I click, it works; the button dissapears from one template, and appears on the other. The second time I click when its on the other template, it doesn't. When compiling, I get this warning: *QMetaObject::connectSlotsByName: No matching signal for on_toolButton_clicked(string)*

Any ideas why the slot stops working?

Answer 1

Any ideas why the slot stops working?

You are missing the signal declaration at the place of the connect as the warning also hints. Also, it seems that you are passing either the slot as the signal to the connect method. A signal should not have the same name as a slot in a Qt application.

Other than that, you may wanna reconsider your design about disabling and enabling a button. Putting them into separate layers is not the appropriate way of doing it.

Moreover, you should probably avoid raw strings for representing states in general. It is better to use enumerations, or boolean for "toggle states".