Question

我的程序中有订单和项目实体类。在order class中有一个Set 项目。当我插入订单时，订单详细信息必须插入订单表。并且必须将Set中的项插入到Items表中，该表在Order表中具有order_id的外键。因此，我想要的查询是：

insert into ordertb (cName) values (?);
insert into item (id, order_id, qty) values(?,?,?);
insert into item (id, order_id, qty) values(?,?,?);
insert into item (id, order_id, qty) values(?,?,?);
// until all the items are inserted

但是hibernate会生成以下查询：

Hibernate: insert into ordertb (cName) values (?)
Hibernate: update item set qty=?, order_id=? where id=?
Hibernate: update item set qty=?, order_id=? where id=?
Hibernate: update item set qty=?, order_id=? where id=?
Hibernate: update item set order_id=? where id=?
Hibernate: update item set order_id=? where id=?
Hibernate: update item set order_id=? where id=?

这是我的映射文件

<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE hibernate-mapping PUBLIC "-//Hibernate/Hibernate Mapping DTD 3.0//EN" "http://hibernate.sourceforge.net/hibernate-mapping-3.0.dtd">
<hibernate-mapping>
    <class name="Order" table="ordertb">
    <meta attribute="class-description">
        This class contains the Order detail. 
    </meta>
    <id name="order" type="int" column="order_id">
        <generator class="native"/>
    </id>
    <property name="cName" column="cName" type="string"/>
    <set name="items" table="item" fetch="select" cascade="all">
        <key>
            <column name="order_id" not-null="true"></column>
        </key>
        <one-to-many class="Item"/>
    </set>
    </class>
</hibernate-mapping>

TestComp.java，我调用方法

import java.util.ArrayList;
import java.util.HashSet;
import java.util.Set;
import org.hibernate.HibernateException;
import org.hibernate.Session;
import org.hibernate.SessionFactory;
import org.hibernate.Transaction;
import org.hibernate.cfg.Configuration;

/*
 * To change this license header, choose License Headers in Project Properties.
 * To change this template file, choose Tools | Templates
 * and open the template in the editor.
 */

/**
 *
 * @author Deepan
 */
public class TestComp {

    private static SessionFactory factory; 

    public static void main(String args[]){
        try{
         factory = new Configuration().configure().buildSessionFactory();
      }catch (Throwable ex) { 
         System.err.println("Failed to create sessionFactory object." + ex);
         System.out.println("Error");
         throw new ExceptionInInitializerError(ex); 

      }
      TestComp ME = new TestComp();

      Order od=new Order();
        Set<Item> items=new HashSet<Item>();

      Item i1=new Item(1, 5, od);
      items.add(i1);
      Item i2=new Item(2, 20, od);
      items.add(i2);
      Item i3=new Item(3, 15, od);
      items.add(i3);

      od.setOrder(200);
      od.setcName("Deepan");
      od.setItems(items);
      /* Add few employee records in database */


      Integer car1 = ME.addOrder(od);
        System.out.println(od.getOrder()+" : "+od.getcName()+" : "+od.getItems());
      //Integer car2 = ME.addOrder(6, "Benz", "gtr");
      //Integer car3 = ME.addOrder(7, "Audi", "sdf");
    }

     public Integer addOrder(Order order) {
        Session session = factory.openSession();
        Transaction tx = null;
        Integer carVIN = null;
        try {
            tx = session.beginTransaction();
            carVIN = (Integer) session.save(order);
            tx.commit();
        } catch (HibernateException e) {
            if (tx != null) {
                tx.rollback();
            }
            e.printStackTrace();
        } finally {
            session.close();
        }
        return carVIN;
    }


}

有什么问题？如果您需要在评论中提及的任何其他信息。由于我对冬眠很新，请给我一个详细的答案。提前谢谢。

Answer 1

将i1，i2，i3作为参数输入方法。然后在carVIN =（Integer）session.save（order）之后添加以下代码;

session.save(i1);
session.save(i2);
session.save(i3);

希望有所帮助

hibernate生成错误的sql查询

1 个答案: