Question

I have a tab widget where every tab is a QTableView. I would like to be able to pass an updated model (QModelIndex) into each tab whenever the contents of that tab need to change.

The alternative (and nastier way) is for me to delete all the tabs, and then recreate them.

I know I can get the widget in the tab by doing something like:

    tabWidget->widget(i);

This will return a widget, which is really a QTableView, but I want to update the model that is in that widget without having to delete and recreate the tab.

Thank you!

P.S. This is my current attempt…

for (int i = 0; i < tableView.size(); i++)
{
    tabWidget->setCurrentWidget(tableView.at(i));

    QTableView* updatedTable = (QTableView*)tabWidget->currentWidget();

    updatedTable->setModel(dataModel);

    tableView.replace(i, updatedTable);
}

Answer 1

It’s not clear why you can’t keep the QTableView widget and just change the model, as in your code. Doesn’t the view refresh without this tableView.replace thing?

There doesn’t appear to be a direct API for replacing the widget you put in with addTab() without going through a tab removal step. But instead of inserting the QTableView directly, you could instead call addTab() on a dummy widget that has a layout in it with a single item. A QStackedLayout, for instance:

QWidget* dummy = new QWidget;
QStackedLayout stackedLayout = new QStackedLayout;
stackedLayout->addWidget(tableView);
dummy->setLayout(stackedLayout);
tabWidget->addTab(dummy);

Then later, when you want to replace the tableView with a new one:

QWidget* dummy = tabWidget->currentWidget();
QStackedLayout newStackedLayout = new QStackedLayout;
newStackedLayout->addWidget(newTableView);
delete dummy->layout();
dummy->setLayout(newStackedLayout);

I still wonder what this is buying you that reusing the old table view couldn’t do.