QTableView - Not Getting Selection Changed Signal

jwomble

I am fairly new to QT, and am having trouble understanding how the QTableView selection changed signal is handled. I have setup a window with an openGL widget and a QTableView. I have a data model class that is correctly populating the tableview, so I added a public slot to that class:

@class APartsTableModel : public QAbstractTableModel
{
public:
AVehicleModel *vehicle;
explicit APartsTableModel(QObject *parent = 0);

//MVC functions
int rowCount(const QModelIndex &parent) const;
int columnCount(const QModelIndex &paret) const;
QVariant data(const QModelIndex &index, int role) const;
QVariant headerData(int section, Qt::Orientation orientation, int role) const;

public slots:
void selectionChangedSlot(const QItemSelection &newSelection,
const QItemSelection &oldSelection);

};@

When I am ready to show the window with the table view, I allocate/initialize it like this:

@//create the display view
AStarModelView *displayWindow = new AStarModelView(this,
starModel->vehicle);

//create the datamodel for the table view
APartsTableModel *dataModel = new APartsTableModel(displayWindow);
dataModel->vehicle = starModel->vehicle;

//create selection model for table view
QItemSelectionModel *selModel = new QItemSelectionModel(dataModel);
displayWindow->materialsTable->setSelectionModel(selModel);

//setup model and signal
displayWindow->materialsTable->setModel(dataModel);

connect(selModel,
SIGNAL(selectionChanged(const QItemSelection &, const QItemSelection &)),
dataModel,
SLOT(selectionChangedSlot(const QItemSelection &, const QItemSelection &)));

//show the view
displayWindow->show();@

When I set a breakpoint in the implementation of the slot function, I never hit it. I've also tried not allocating a new QItemSelectionModel, but that didn't work either. I'm really not sure what I'm doing wrong here.

daviddoria

(There indeed is no selectionChanged signal of a QTableView, sorry)
-Do you really need the QItemSelectionModel? It doesn't sound like you are doing anything that complex. I think you want to connect the selectionChanged model of the VIEW to a slot.-

-connect(materialsTable,-
-SIGNAL(selectionChanged(const QItemSelection &, const QItemSelection &)),-
-dataModel,-
-SLOT(selectionChangedSlot(const QItemSelection &, const QItemSelection &)));-

I would also name your materialsTable materialsTableView or something like that - as to not confuse if the 'table' is a 'tableModel' or a 'tableView'.

jwomble

Does QTableView have a selectionChanged signal? I can't find it in the class reference.

daviddoria

Yep, it's in the superclass: http://developer.qt.nokia.com/doc/qt-4.8/qabstractitemview.html#signalSection

I was just complaining about that here :) : http://developer.qt.nokia.com/forums/viewthread/12604/

jwomble

I see a slot for selectionChanged() in QAbstractItemView, but not a signal.

daviddoria

Yikes - please ignore everything I've said so far - I was a bit confused (clearly).

Can you do this - handle the clicked() signal of the view (or the mouseRelease event if you're making bigger selections) and then use the view's selectedIndexes function to get the selection?

daviddoria

Gah - it is listed under public functions (http://developer.qt.nokia.com/doc/qt-4.8/qlistview.html#memberSection) even though it is protected (it even says protected right in the declaration in the doc!). Sorry, without subclassing the view this method won't do it for you.

jwomble

As far as I can tell, using the selectionModel is the correct way to do this, I just can't figure out what I'm doing wrong.

jwomble

What's the first thing you should check in QT when signals/slots don't seem to be working correctly? That your class has the Q_OBJECT macro in it. Added this to the APartsTable class definition, and now I'm hitting the breakpoint.

When does Friday get here?

daviddoria

Here is a compilable example:

http://programmingexamples.net/wiki/Qt/ModelView/ItemSelectionModel

goetz

The table implicitly has a selction model, get with <code>tableView->selectionModel()</code>, you don't need to create a new one.