Accessing specific controls from widgets

jsulm

@IknowQT said in Accessing specific controls from widgets:

You want to access the PchkItem inside the pChkWidget.

I already wrote you that you need to cast QWidget* to the type you need.
Did you actually read my response?

IknowQT

for (int i = 0; i < this->ui.fileTableWidget->rowCount(); i++)
{
usrCheckBox* pChkItem = static_cast<usrCheckBox*>(this->ui.fileTableWidget->cellWidget(i, 0));
qDebug() << "GetSelectedItems()"<< pChkItem->GetCheckID();
}

I tested it like this.
I didn't get the results I wanted.

Casting the widget doesn't seem to work.

jsulm

@IknowQT You should really read more carefully! I suggested to use https://doc.qt.io/qt-6/qobject.html#qobject_cast-1 for casting, not static_cast.

IknowQT

@jsulm

usrCheckBox* pChkItem = qobject_cast<usrCheckBox*>(this->ui.fileTableWidget->cellWidget(i, 0));

I tested it, but it comes out as Null after casting.
I don't think there is any way to access this

ChrisW67

Yes, that is because the cell widget is not of type usrCheckBox.

jsulm

@IknowQT said in Accessing specific controls from widgets:

I don't think there is any way to access this

Of course there is, but as @ChrisW67 wrote: what this->ui.fileTableWidget->cellWidget(i, 0) returns is apparently not an usrCheckBox...

IknowQT

@jsulm

QWidget* pChkWidget = new QWidget();

If I do setCellWidget after declaring it as usrCheckBox without declaring it as qwidget, it seems to come out as I want.

But I have to use Layout. please see my code

JonB

@IknowQT
Please be logical about this, and examine your own code. So far as I can see, glancing, I think you have a hierarchy like:

setCellWidget(nRowIndex, 0, pChkWidget);
    QWidget* pChkWidget = new QWidget()
        QHBoxLayout* layout = new QHBoxLayout(pChkWidget);
            usrCheckBox* pChkItem = new usrCheckBox(pChkWidget);

So clearly picking up the cellWidget() is not going to return your usrCheckBox, it is going to return your pChkWidget, right?

So from there to get to the usrCheckBox you must start from that widget and then use one of:

Find its child QHBoxLayout, find the child QWidget which is on that, and cast that to usrCheckBox*; or
Call QObject::findChild<usrCheckBox*>() on it to let it find the (first) descendant which is of type usrCheckBox*.

jsulm

@IknowQT So, in fact you're setting pChkWidget (which is a QWidget) as cell widget, right? Then why do you expect this->ui.fileTableWidget->cellWidget(i, 0) to return usrCheckBox?!
As @JonB wrote usrCheckBox is a child of pChkWidget. So get pChkWidget first (calling this->ui.fileTableWidget->cellWidget(i, 0)) and then its child usrCheckBox.

ChrisW67

Another approach would be to define your own widget class that exposes the state of the check box it contains.

class FancyCellWidget: pubic QWidget {
  Q_OBJECT
public:
  explicit FancyCellWidget(QWidget *p = nullptr);
  Qt::CheckState 	checkState() const; // just return this from the internal QCheckbox
  void 	setCheckState(Qt::CheckState state); // pass through to the internal QCheckbox
  ...
signals:
  void clicked(); // connect the internal check box clicked() signal to this one in the constructor
};

and then use that to create your cell widget.

IknowQT

@jsulm

@jsulm said in Accessing specific controls from widgets:

Yes. What exactly is the problem?
I think you need to cast from QWidget* to what ever the widget actually is (https://doc.qt.io/qt-6/qobject.html#qobject_cast-1).

@IknowQT said in Accessing specific controls from widgets:

Inside the widget of pChkWidget is the pChkItem control.
The return value of this->ui.fileTableWidget->cellWidget(row, col) is Widget.
You want to access the PchkItem inside the pChkWidget.

You recommended objectcast.
I said clearly. I have a Checkbox control inside a Widget, so if I cast an object, does it become a checkbox?

I showed the test results. Returning as null is a natural result.
I think you suggested just object cast without looking closely at the code you asked in the first place.

@JonB
I was able to access that control using findChild, which Jonby recommended.

jsulm

@IknowQT said in Accessing specific controls from widgets:

so if I cast an object, does it become a checkbox

Casting does not magically change the type of an object. That's why qobject_cast will return nullptr if the object you're trying to cast is not of the type you want to cast to. Casting only works if the object is really an object of the target type.