Question

我正在尝试在ListView上显示所选ID的所有记录，但是如何在Hibernate中显示来自“name”列中所选ID的记录？在MySQL中它看起来像这样，它工作正常：

@Override
        public void changed(ObservableValue<? extends String> observable, String oldValue, String idvalue) {
            try {
            pst = con.prepareStatement("SELECT * FROM przychodniadb.patient WHERE idpatient=" + idvalue);
                rs = pst.executeQuery();

                while (rs.next()) {
                    infoPIDField.setText(rs.getString(1));
                    infonameField.setText(rs.getString(2));
                    infolastNameField.setText(rs.getString(3));
                    infogenderField.setText(rs.getString(4));
                    infoageField.setText(rs.getString(5));
                    infophonenumberField.setText(rs.getString(6));
                    infoadressField.setText(rs.getString(7));
                    infodiseaseField.setText(rs.getString(8));
                    infoconditionField.setText(rs.getString(9));
                    infodataField.setText(rs.getString(12));
                    infoRoomNumber.setText(rs.getString(13));
                    infoRoomType.setText(rs.getString(10));*/

                }

            } catch (SQLException e) {

                e.printStackTrace();
            }

        }
    });

现在在Hibernate：

@Override
        public void changed(ObservableValue<? extends String>  observable, String oldValue, String idvalue) {

Query query3 = entityManager.createQuery("FROM Patient WHERE idpatient=:idpatient");
                query3.setParameter("idpatient", idvalue);


                    infoPIDField.setText(???);
                    infonameField.setText(???);
                    infolastNameField.setText(???);
                    ...........


    });

我应该把什么作为参数？

Answer 1

首先在Patient对象中取结果：

Patient c = (Patient) query.getSingleResult();

此后，您可以使用简单的getter访问这些字段。

@Override
public void changed(ObservableValue<? extends String>  observable, String oldValue, String idvalue) {

    Query query3 = entityManager.createQuery("FROM Patient WHERE idpatient=:idpatient");
    query3.setParameter("idpatient", idvalue);

    Patient p = (Patient) query.getSingleResult();


    infoPIDField.setText(p.getPIDField());
    infonameField.setText(p.getNameField());
    infolastNameField.setText(p.getLastNameField());
    ...........


}

JavaFX中的Hibernate - 显示所选ID的记录

1 个答案: