scala – 在SAP HANA中选择Hibernate抛出错误：`不支持com.sap.d

CREATE column TABLE banks (
  sk tinyint NOT NULL GENERATED BY DEFAULT AS IDENTITY,code varchar(10) DEFAULT NULL,name varchar(100) DEFAULT NULL,version smallint DEFAULT NULL,PRIMARY KEY (sk)
);

我尝试使用以下代码选择表的行(在Scala中)：

import scala.collection.JavaConverters._

object Test extends App {

    val session = HibernateUtil.sessionFactory.openSession        
    val q = session.createQuery("from BankHib ") 
    val list2 = q.list   // <-- code breaks here

    session.close
 }

使用以下实体定义：

@Entity
@Table(name = "banks")
class BankHib {

    @Id
    var sk: Int = _

    var code: String = _
    var name: String = _
    var version: Int = _
}

以及获取会话工厂的实用程序：

object HibernateUtil {

  val sessionFactory = buildSessionFactory

    def buildSessionFactory = {
        try {
                new Configuration().configure().buildSessionFactory();
        } catch {case ex:Throwable => 
            println("Initial SessionFactory creation failed." + ex);
            throw new ExceptionInInitializerError(ex);
        }
    }

    def shutdown  {
        sessionFactory.close
    } 
}

当我运行Test对象时,我得到以下异常：

Caused by: com.sap.db.jdbc.exceptions.SQLFeatureNotSupportedExceptionSapDB: Method unwrap of com.sap.db.jdbc.CallableStatementSapDBFinalize is not supported.
    at com.sap.db.jdbc.exceptions.SQLExceptionSapDB._createException(SQLExceptionSapDB.java:155)
    at com.sap.db.jdbc.exceptions.SQLExceptionSapDB.generateSQLException(SQLExceptionSapDB.java:26)
    at com.sap.db.jdbc.WrapperDummy.unwrap(WrapperDummy.java:25)
    at org.hibernate.engine.jdbc.internal.ResultSetReturnImpl.extract(ResultSetReturnImpl.java:64)
    ... 26 more

有什么问题以及如何解决？什么是不支持的功能？