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QSqlRelationalTableModel && removeRow() [Solved]

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  • L Offline
    L Offline
    luca
    wrote on last edited by
    #3

    [quote author="kunashir" date="1310101354"]Is Select() work fine? [/quote]

    Yes, I get a populated QTableVew.

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    • L Offline
      L Offline
      luca
      wrote on last edited by
      #4

      I found some discussion where someone has the same problem: "this":http://bugreports.qt.nokia.com/browse/QTBUG-5027 and "this":http://www.developer.nokia.com/Community/Discussion/showthread.php?190187-QSqlRelationalTableModel-removeRow-and-QSqlRelationalTableModel-removeRows-not-work . But no one post a solution...

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      • K Offline
        K Offline
        kunashir
        wrote on last edited by
        #5

        I was used it example - and all work fine:

        @ #include <QtGui>
        #include <QtSql>

        int main(int argc, char *argv[]) {
        QApplication app(argc, argv);
        QTextCodec *codec = QTextCodec::codecForName("CP1251");
        QTextCodec::setCodecForTr(codec);

        QSqlDatabase db = QSqlDatabase::addDatabase("QSQLITE");
        db.setDatabaseName("mydb1");
        db.setUserName("...");
        db.setPassword("...");
        db.open();
        
        // create and fill table
        QSqlQuery query;
        query.exec&#40;"create table employee( "
                   " id int primary key, "
                   " name varchar(20&#41;, "
                   " department int, "
                   " position int&#41;" );
        query.exec&#40;QObject::tr("insert into employee values(1, 'Ivanov A.K.', 1, 3&#41;"&#41;);
        query.exec&#40;QObject::tr("insert into employee values(2, 'Petorv I.E.', 3, 2&#41;"&#41;);
        query.exec&#40;QObject::tr("insert into employee values(3, 'Mihailov N.P.', 2, 1&#41;"&#41;);
        
        query.exec&#40;"create table department(id int, name varchar(20&#41;&#41;");
        query.exec&#40;QObject::tr("insert into department values(1, 'Administration'&#41;"&#41;);
        query.exec&#40;QObject::tr("insert into department values(2, 'Accounting'&#41;"&#41;);
        query.exec&#40;QObject::tr("insert into department values(3, 'Planning Department&#41;"&#41;);
        
        query.exec&#40;"create table position(id int, name varchar(20&#41;&#41;");
        query.exec&#40;QObject::tr("insert into position values(1, 'Operator'&#41;"&#41;);
        query.exec&#40;QObject::tr("insert into position values(2, 'Economist'&#41;"&#41;);
        query.exec&#40;QObject::tr("insert into position values(3, 'Boss'&#41;"&#41;);
        
        QSqlRelationalTableModel model;
        model.setTable("employee");
        model.setSort(1, Qt::AscendingOrder);
        
        model.setEditStrategy(QSqlTableModel::OnFieldchange);//but this other stategy
        
        // set relation
        model.setRelation(2, QSqlRelation("department", "id", "name"));
        model.setRelation(3, QSqlRelation("position", "id", "name"));
        
        model.setHeaderData(0, Qt::Horizontal, QObject::tr("Number"));
        model.setHeaderData(1, Qt::Horizontal, QObject::tr("Name"));
        model.setHeaderData(2, Qt::Horizontal, QObject::tr("department"));
        model.setHeaderData(3, Qt::Horizontal, QObject::tr("Post"));
        
        model.select();
        
        
        // create view
        QTableView view;
        view.setModel(&model);
        
        // create delegate
        view.setItemDelegate(new QSqlRelationalDelegate(&view));
        
        view.setWindowTitle(QObject::tr("Relation table"));
        view.show();
        model.removeRow(1);
        return app.exec&#40;&#41;;
        

        }@

        Can You show scheme of DB?

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        • L Offline
          L Offline
          luca
          wrote on last edited by
          #6

          I have this situation:
          @
          CREATE TABLE vs_applicazioni (
          id int(11) NOT NULL,
          nome varchar(100) NOT NULL,
          descrizione varchar(250) NOT NULL,
          menu varchar(20) NOT NULL,
          funzione_di_avvio varchar(200) DEFAULT NULL,
          hidden tinyint(1) DEFAULT '0',
          PRIMARY KEY (id),
          KEY menu (menu),
          CONSTRAINT vs_applicazioni_ibfk_1 FOREIGN KEY (menu) REFERENCES vs_menu (nome)
          )

          CREATE TABLE vs_permessi (
          id_gruppo int(11) NOT NULL,
          id_applicazione int(11) NOT NULL,
          PRIMARY KEY (id_gruppo,id_applicazione),
          KEY id_applicazione (id_applicazione),
          CONSTRAINT vs_permessi_ibfk_1 FOREIGN KEY (id_applicazione) REFERENCES vs_applicazioni (id),
          CONSTRAINT vs_permessi_ibfk_2 FOREIGN KEY (id_gruppo) REFERENCES vs_gruppi_utenti (id)
          )
          @

          And I'm showing in a QTableView the vs_permessi table replacing id_applicazione column with vs_applicazioni.descrizione column . It show the correct data.

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          • K Offline
            K Offline
            kunashir
            wrote on last edited by
            #7

            I suppose, that the complex primary key for QSqlRelationalTableModel is a problem.
            Maybe, easier to create your own model?

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            • L Offline
              L Offline
              luca
              wrote on last edited by
              #8

              [quote author="kunashir" date="1310112233"]I suppose, that the complex primary key for QSqlRelationalTableModel is a problem.
              Maybe, easier to create your own model?[/quote]

              I also tried with this tables:
              @
              CREATE TABLE vs_applicazioni (
              id int(11) NOT NULL,
              nome varchar(100) NOT NULL,
              descrizione varchar(250) NOT NULL,
              menu varchar(20) NOT NULL,
              funzione_di_avvio varchar(200) DEFAULT NULL,
              hidden tinyint(1) DEFAULT '0',
              PRIMARY KEY (id),
              KEY menu (menu)
              )

              CREATE TABLE vs_permessi (
              id_gruppo int(11) NOT NULL,
              id_applicazione int(11) NOT NULL,
              PRIMARY KEY (id_gruppo,id_applicazione),
              KEY id_applicazione (id_applicazione)
              )
              @

              with the same result...

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              • L Offline
                L Offline
                luca
                wrote on last edited by
                #9

                I also tried your example with my table (the versione without foregn key):
                @
                #include <QtGui>
                #include <QtSql>

                int main(int argc, char *argv[]) {
                QApplication app(argc, argv);
                QTextCodec *codec = QTextCodec::codecForName("CP1251");
                QTextCodec::setCodecForTr(codec);

                QSqlDatabase db = QSqlDatabase::addDatabase("QMYSQL");
                db.setDatabaseName("vs_manager");
                db.setUserName("luca");
                db.setPassword("luca123");
                
                if(!db.open())
                {
                    qDebug() << db.lastError().number();
                    
                }
                
                QSqlRelationalTableModel model;
                model.setTable("vs_permessi");
                model.setSort(1, Qt::AscendingOrder);
                
                model.setEditStrategy(QSqlTableModel::OnFieldChange);//but this other stategy
                
                // set relation
                model.setRelation(1,QSqlRelation("vs_applicazioni","id","descrizione"));
                
                model.select();
                
                
                // create view
                QTableView view;
                view.setModel(&model);
                
                // create delegate
                view.setItemDelegate(new QSqlRelationalDelegate(&view));
                
                view.setWindowTitle(QObject::tr("Relation table"));
                view.show();
                model.removeRow(1);
                return app.exec();
                

                }

                @

                but I get the same error:
                @
                QSqlQuery::value: not positioned on a valid record
                @

                Should it be a mysql driver problem?

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                • K Offline
                  K Offline
                  kunashir
                  wrote on last edited by
                  #10

                  I don't understand why:
                  @
                  m_applicazioniTableModel->setRelation(1,QSqlRelation("vs_applicazioni","id","descrizione"));
                  @

                  at the same time, table vs_permessi don't have column "descrizione".
                  If I rightly understand Your scheme.

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                  • L Offline
                    L Offline
                    luca
                    wrote on last edited by
                    #11

                    this should mean that I want "replace" the column 1 of vs_permessi table with the column descrizione of vs_applicazioni table linking it with vs_applicazioni.id .
                    The table view show the correct data.

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                    • L Offline
                      L Offline
                      luca
                      wrote on last edited by
                      #12

                      Now I tried to compile the example in Windows using ODBC to connect to the MySql DB with the same result...

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                      • K Offline
                        K Offline
                        kunashir
                        wrote on last edited by
                        #13

                        Hmm....
                        Query of removed is not correct
                        "DELETE FROM vs_permessi WHERE id_gruppo = ? AND id_applicazione IS NULL"
                        but I'm don't know why....
                        Sorry, his have also some value...

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                        • L Offline
                          L Offline
                          luca
                          wrote on last edited by
                          #14

                          I found the cause.
                          The table vs_permessi:
                          @
                          CREATE TABLE vs_permessi (
                          id_gruppo int(11) NOT NULL,
                          id_applicazione int(11) NOT NULL,
                          PRIMARY KEY (id_gruppo,id_applicazione),
                          KEY id_applicazione (id_applicazione)
                          )
                          @
                          has the column id_applicazione in its primary key and this seems the problem.

                          If I modify the table this way:
                          @
                          CREATE TABLE vs_permessi (
                          id_gruppo int(11) NOT NULL,
                          id_applicazione int(11) NOT NULL,
                          PRIMARY KEY (id_gruppo)
                          )
                          @

                          the problem disappears...

                          Very thanks for you time and your help...

                          1 Reply Last reply
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