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我试图显示所选择的ID在ListView中的所有记录,但我怎么能显示例如记录从选择ID从列'名称'在休眠?在MySQL中,它看起来像它工作正常:在JavaFX中休眠 - 显示选定ID的记录
@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();
}
}
});
现在休眠:
@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(???);
...........
});
我应该把作为参数?