QSqlRelationalTableModel doesn't give relation foreign key with Qt::EditRole

[BrokenVoodooDoll]

Consider the Qt's example "RelationalTableModel".

#include <QtWidgets>
#include <QtSql>

#include "../connection.h"
#include <QDebug>

#include <memory>

void initializeModel(QSqlRelationalTableModel *model)
{
//! [0]
    model->setTable("employee");
//! [0]

    model->setEditStrategy(QSqlTableModel::OnManualSubmit);
//! [1]
    model->setRelation(2, QSqlRelation("city", "id", "name"));
//! [1] //! [2]
    model->setRelation(3, QSqlRelation("country", "id", "name"));
//! [2]

//! [3]
    model->setHeaderData(0, Qt::Horizontal, QObject::tr("ID"));
    model->setHeaderData(1, Qt::Horizontal, QObject::tr("Name"));
    model->setHeaderData(2, Qt::Horizontal, QObject::tr("City"));
    model->setHeaderData(3, Qt::Horizontal, QObject::tr("Country"));
//! [3]

    model->select();
}

std::unique_ptr<QTableView> createView(const QString &title, QSqlTableModel *model)
{
//! [4]
    std::unique_ptr<QTableView> view{new QTableView};
    view->setModel(model);
    view->setItemDelegate(new QSqlRelationalDelegate(view.get()));
//! [4]
    view->setWindowTitle(title);

    QObject::connect(view.get()->selectionModel(), &QItemSelectionModel::selectionChanged,
                     [=](const QItemSelection &selected, const QItemSelection &deselected){
        for (const QModelIndex &idx : selected.indexes()) {
            qDebug() << "DisplayRole:" << model->index(idx.row(), idx.column(), QModelIndex()).data(Qt::DisplayRole);
            qDebug() << "EditRole" << model->index(idx.row(), idx.column(), QModelIndex()).data(Qt::EditRole);
        }
        qDebug() << '\n';
    });

    return view;
}

void createRelationalTables()
{
    QSqlQuery query;
    query.exec("create table employee(id int primary key, name varchar(20), city int, country int)");
    query.exec("insert into employee values(1, 'Espen', 5000, 47)");
    query.exec("insert into employee values(2, 'Harald', 80000, 49)");
    query.exec("insert into employee values(3, 'Sam', 100, 1)");

    query.exec("create table city(id int, name varchar(20))");
    query.exec("insert into city values(100, 'San Jose')");
    query.exec("insert into city values(5000, 'Oslo')");
    query.exec("insert into city values(80000, 'Munich')");

    query.exec("create table country(id int, name varchar(20))");
    query.exec("insert into country values(1, 'USA')");
    query.exec("insert into country values(47, 'Norway')");
    query.exec("insert into country values(49, 'Germany')");
}

int main(int argc, char *argv[])
{
    QApplication app(argc, argv);
    if (!createConnection())
        return EXIT_FAILURE;

    createRelationalTables();

    QSqlRelationalTableModel model;

    initializeModel(&model);

    std::unique_ptr<QTableView> view = createView(QObject::tr("Relational Table Model"), &model);
    view->show();

    return app.exec();
}

I've added some debug code into the "createView" function.
When I run the application and select, for example, the cell on the first row and the "City" column in the table, I get the following message in the console:
DisplayRole: QVariant(QString, "Oslo")
EditRole QVariant(QString, "Oslo")
Then I double-click that cell and in the combo-box, I change the city to "San Jose", for example, and again select this cell. Now I get another message in the console:
DisplayRole: QVariant(QString, "San Jose")
EditRole QVariant(qlonglong, 100).
As you can see, now the "EditRole" flag allows us to get the ID of a city in the relational table. But one cannot obtain that ID not changing the value in the cell.
How can I get that ID at once, not changing manually values in such cells?

[SGaist]

Hi,

You can use the model returned by QRelationalTableModel::relationModel to get the information you seek.