Question

我能够使用hibernate在关系数据库中持久化对象。请看下面的代码。

package one;

import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.GeneratedValue;
import javax.persistence.Id;
import javax.persistence.SecondaryTable;
import javax.persistence.Table;

@Entity
public class Customer {

    @Id
    private int customerId;
    private String customerName;
    private String customerAddress;
    private int creditScore;
    private int rewardPoints;

    public Customer()
    {

    }

    public Customer(int customerId,String customerName,String customerAddress,int creditScore,int rewardsPoints)
    {
        this.customerId=customerId;
        this.customerAddress=customerAddress;
        this.creditScore=creditScore;
        this.customerName=customerName;
        this.rewardPoints=rewardsPoints;
    }

    public int getCustomerId() {
        return customerId;
    }
    public void setCustomerId(int customerId) {
        this.customerId = customerId;
    }
    public String getCustomerName() {
        return customerName;
    }
    public void setCustomerName(String customerName) {
        this.customerName = customerName;
    }


    public String getCustomerAddress() {
        return customerAddress;
    }
    public void setCustomerAddress(String customerAddress) {
        this.customerAddress = customerAddress;
    }


    public int getCreditScore() {
        return creditScore;
    }
    public void setCreditScore(int creditScore) {
        this.creditScore = creditScore;
    }

    public int getRewardPoints() {
        return rewardPoints;
    }
    public void setRewardPoints(int rewardPoints) {
        this.rewardPoints = rewardPoints;
    }

}

然后保存这个类的对象我使用了下面的类。下面的类创建Customer类的对象并将该对象保存在数据库中，然后再次检索它并打印每个已保存对象的CustomerName属性。

package one;

import java.util.ArrayList;
import java.util.Iterator;
import java.util.List;

import org.hibernate.Session;
import org.hibernate.SessionFactory;
import org.hibernate.cfg.Configuration;

public class TestCustomer {
    public static void main(String args[])
    {
        Customer cust = new Customer(13,"Sara","Banglore",9000,60);


        SessionFactory factory = new Configuration().configure().buildSessionFactory();
        Session session = factory.openSession();
        session.beginTransaction();
        session.save(cust);
        session.getTransaction().commit();
        session.close();

        session = factory.openSession();
        session.beginTransaction();
        List list = session.createQuery("FROM Customer").list();
        Iterator iterator = list.iterator();

        while(iterator.hasNext())
        {
            Customer custA = (Customer)iterator.next();
            System.out.println("First Name\t"+custA.getCustomerName());
        }
        session.getTransaction().commit();
        session.close();

    }
}

我在代码上执行了很多次。代码运行正常。它能够获取所有保存的对象。但后来我使用了oracle toad并发出sql语句作为

Insert into Customer(CUSTOMERID,CREDITSCORE,CUSTOMERNAME,REWARDPOINTS,CUSTOMERADDRESS)
VALUES(87,4000,'Saurabh',20,'Kalwa');

记录存储在表中但是当我执行上面的代码时，我无法获取此记录。我可以得出的一个结论是hibernate只返回persisted objects，但是还有其他方法可以获得所有记录吗？

Answer 1

您是否确定在使用toad插入oracle后提交了记录？（您可以打开另一个客户端并执行select以确保它可以从sql客户端获取）。
如果要调试，可以启用hibernate的sql日志功能，然后在sql客户端执行hibernate为查询生成的sql，以确保可以正确获取所有记录。

使用JPA的一些建议：

确保@Entity具有映射到物理表的名称值，以避免表映射混淆。
使用@Column（name =＆＃34; column＆＃34;）将所有字段映射到物理表列，以避免混淆。

使用hibernate从表中获取数据

1 个答案: