在Hibernate中插入多行 - 批量插入

Question

我在我的DAO类中使用以下事务代码单元在Hibernate中插入多行.

@Override
@SuppressWarnings("unchecked")
@Transactional(readOnly = false, propagation=Propagation.REQUIRES_NEW, rollbackFor={Throwable.class})
public String insertZoneCharge(Long zoneId, List<Object[]> items)
{
    Session session=sessionFactory.openSession();
    session.beginTransaction();

    //Configuration configuration=new Configuration();
    //configuration.setProperty("hibernate.jdbc.batch_size", "50");
    int i=1;

    for(Object[] o:items)
    {
        if(o[2]!=null&&StringUtils.isNotBlank(o[2].toString()))
        {
            ZoneChargePK zoneChargeId=new ZoneChargePK();
            zoneChargeId.setWeightId(Long.parseLong(o[0].toString()));
            zoneChargeId.setZoneId(zoneId);

            model.ZoneCharge zoneCharge=new model.ZoneCharge();
            zoneCharge.setZoneChargePK(zoneChargeId);
            zoneCharge.setCharge(new BigDecimal(o[2].toString()).setScale(2, RoundingMode.HALF_UP));
            session.save(zoneCharge);

            if(++i%50==0)
            {
                session.flush();
                session.clear();
            }
        }
    }

    session.getTransaction().commit();
    session.close();
    return "Data saved successfully.";
}

此方法的参数通过JSON请求提供.我不会深入了解Oracle表的详细信息,因为它不相关.

当我提供让我们说3个值时,这个循环被迭代三次并且在相应的数据库表中创建了三行.成功完成事务后,INSERT我在控制台上看到的以下本机Oracle 语句似乎被执行了三次.

insert into WAGAFASHIONDB.ZONE_CHARGE (CHARGE, WEIGHT_ID, ZONE_ID) values (?, ?, ?)

调试和跟踪信息如下.

DEBUG [http-apr-8080-exec-84] (SqlStatementLogger.java:104) - insert into WAGAFASHIONDB.ZONE_CHARGE (CHARGE, WEIGHT_ID, ZONE_ID) values (?, ?, ?)
TRACE [http-apr-8080-exec-84] (BasicBinder.java:83) - binding parameter [1] as [NUMERIC] - 1.00
TRACE [http-apr-8080-exec-84] (BasicBinder.java:83) - binding parameter [2] as [BIGINT] - 164
TRACE [http-apr-8080-exec-84] (BasicBinder.java:83) - binding parameter [3] as [BIGINT] - 22

DEBUG [http-apr-8080-exec-84] (SqlStatementLogger.java:104) - insert into WAGAFASHIONDB.ZONE_CHARGE (CHARGE, WEIGHT_ID, ZONE_ID) values (?, ?, ?)
TRACE [http-apr-8080-exec-84] (BasicBinder.java:83) - binding parameter [1] as [NUMERIC] - 2.00
TRACE [http-apr-8080-exec-84] (BasicBinder.java:83) - binding parameter [2] as [BIGINT] - 221
TRACE [http-apr-8080-exec-84] (BasicBinder.java:83) - binding parameter [3] as [BIGINT] - 22

DEBUG [http-apr-8080-exec-84] (SqlStatementLogger.java:104) - insert into WAGAFASHIONDB.ZONE_CHARGE (CHARGE, WEIGHT_ID, ZONE_ID) values (?, ?, ?)
TRACE [http-apr-8080-exec-84] (BasicBinder.java:83) - binding parameter [1] as [NUMERIC] - 3.00
TRACE [http-apr-8080-exec-84] (BasicBinder.java:83) - binding parameter [2] as [BIGINT] - 163
TRACE [http-apr-8080-exec-84] (BasicBinder.java:83) - binding parameter [3] as [BIGINT] - 22

因此,似乎INSERT已经执行了三个单独的语句,并且已经进行了三次单独的数据库访问.

我希望它INSERT在一次旅行中只执行一个多行语句.在这种情况下,Hibernate是否真的执行了三个单独的语句？如果是,那么在特定批次中执行语句的方式是什么？我的代码有什么问题吗？

---

我的Hibernate配置application-context.xml如下.

<property name="hibernateProperties">
    <value>
        hibernate.dialect=org.hibernate.dialect.Oracle10gDialect
        hibernate.jdbc.batch_size=50
        hibernate.cache.provider_class=org.hibernate.cache.EhCacheProvider            
        hibernate.format_sql=false
        hibernate.show_sql=true
        hibernate.order_updates=true
        hibernate.connection.autocommit=false
        hibernate.order_inserts=true
        hibernate.order_updates=true
        hibernate.cache.use_query_cache=false
        hibernate.cache.use_second_level_cache=false
        javax.persistence.validation.mode=callback            
    </value>        
  </property>           

Answer 1

你的期望是错误的.Hibernate不会生成单个insert语句.但它会为批处理添加三个插入语句并执行此批处理,从而对数据库进行单次往返.

有关JDBC批处理更新的更多信息,请阅读http://docs.oracle.com/javase/1.3/docs/guide/jdbc/spec2/jdbc2.1.frame6.html.