QSqlQuery error

gabor53

Hi,
I've spent quite some time trying to figure out what's wrong with this code. Whenever I run it I get the following messages:
The database is open!
QSqlQuery::value: not positioned on a valid record

The code:

    QSqlDatabase db;
    db = QSqlDatabase::addDatabase ("QSQLITE");
    db.setDatabaseName ("C:/Programming/Qtsamples/Image_from_DB/db.db");
	db.open ();

    QSqlQuery query;

    if(!db.open ())
    {
        qDebug() << "The database is NOT open!";
    }
    else
    {
        qDebug() << "The database is open!";
    }

    query.prepare ("SELECT Pic FROM Items");
    query.exec ();



    query.first ();
	QByteArray ByteArray;
	ByteArray = query.value (1).toByteArray ();
    QPixmap Pixmap = QPixmap();
    Pixmap.loadFromData (ByteArray);
    db.close ();

	ui->label->setPixmap (Pixmap);
    ui->label->show ();

Please help me to make it positioned on a valid record!
Thank you.

kshegunov

@gabor53
Hello,
You select one column from the table, but try to retrieve the second column from the resultset, try:

    query.value(0).toByteArray();

Additional (potential) problems exist in your code:

In example, you don't handle the return value of your QSqlDatabase::first call, it's supposed to tell you whether you're properly positioned at the first record or not.
You call QSqlDatabase::open twice, you can use QSqlDatabase::isOpen to check if your database was properly opened instead of trying to open it again.
You don't handle the return value of QSqlQuery::exec as well.
There is no need to really prepare the query before execution, you don't intend to use it multiple times with different bindings.
You close your database after executing the query. Usually the idea is to open/close the database once and have multiple queries executed on the same database instance.

Kind regards.

gabor53

Thank you. it worked.