Hibernate @Version 注释

hibernate @version 和 ManyToOne Mapping 之间的关系是什么？

假设我有两个表“部门”和“员工”。这是部门是主表 和明细表中的员工。在 Employee 表中，部门 ID 作为外键引用。

这是我的课程

Public class Department {

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private long ID;
    @Version
    private Long version;

    //Getters and Setters

}

public class Employee {

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private long ID;
    @Version
    private Long version;

    @ManyToOne(fetch = FetchType.LAZY)
    @JoinColumn(name = "departmentID" )
    private Department department;

}

而且，Spring 处理会话。因此，假设在一页中获取并存储了特定部门 在 HTTP 会话中。

现在在另一个页面中，我正在尝试执行以下操作

Employee emp = new Employee();
emp.setName('Test')
emp.setDepartment(dept) // already stored in the HTTP session variable
service.save(emp)

现在我收到以下异常

org.springframework.dao.InvalidDataAccessApiUsageException: object references an unsaved transient instance - save the transient instance before flushing: 

只需进行如下一处更改即可出现错误

Employee emp = new Employee();
emp.setName('Test')
dept.setVersion(0);
emp.setDepartment(dept) // already stored in the HTTP session variable
service.save(emp)

我的 Spring 配置

<?xml version="1.0" encoding="UTF-8"?>
<beans xmlns="http://www.springframework.org/schema/beans"
xmlns:p="http://www.springframework.org/schema/p" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns:context="http://www.springframework.org/schema/context" xmlns:tx="http://www.springframework.org/schema/tx"
xsi:schemaLocation="
       http://www.springframework.org/schema/beans
       http://www.springframework.org/schema/beans/spring-beans-3.0.xsd
       http://www.springframework.org/schema/context
       http://www.springframework.org/schema/context/spring-context-3.0.xsd
       http://www.springframework.org/schema/tx
               http://www.springframework.org/schema/tx/spring-tx-3.0.xsd">

<bean id="transactionManager"
    class="org.springframework.orm.hibernate4.HibernateTransactionManager">
    <property name="sessionFactory" ref="sessionFactory" />
</bean>

<!-- Container Configuration: The IOC container configuration xml file is 
    shown below,The container has the <context:component-scan> element and <context:annotation-config/> 
    <context:annotation-config/> used to intimate the beans of this IOC container 
    are annotation supported. By pass the base path of the beans as the value 
    of the base-package attribute of context:component-scan element, we can detect 
    the beans and registering their bean definitions automatically without lots 
    of overhead. The value of base-package attribute is fully qualified package 
    name of the bean classes. We can pass more than one package names by comma 
    separated -->

<context:annotation-config />
<context:component-scan base-package="com.product.business" />

<tx:annotation-driven transaction-manager="transactionManager" />

<!-- This will ensure that hibernate or jpa exceptions are automatically 
    translated into Spring's generic DataAccessException hierarchy for those 
    classes annotated with Repository -->

<bean
    class="org.springframework.dao.annotation.PersistenceExceptionTranslationPostProcessor" />

<bean id="CRUDService" class="com.product.business.service.CRUDServiceImpl" />
<bean id="AuthService" class="com.product.business.service.AuthServiceImpl" />

服务实施

package com.product.business.service;

import java.io.Serializable;
import java.util.List;

import org.springframework.beans.factory.annotation.Autowired;
import org.springframework.dao.DataAccessException;
import org.springframework.stereotype.Service;
import org.springframework.transaction.annotation.Transactional;

import com.product.business.dao.CRUDDao;

@Service
public class CRUDServiceImpl implements CRUDService {

    @Autowired
    private CRUDDao CRUDDao;

    @Transactional(readOnly = true)
    public <T> List<T> getAll(Class<T> klass) {
        return CRUDDao.getAll(klass);
    }

    @Transactional
    public <T> void Save(T klass) throws DataAccessException {
        CRUDDao.Save(klass);
    }

    @Transactional
    public <T> void delete(T klass) throws DataAccessException {
        CRUDDao.delete(klass);
    }

    @Transactional
    public <T> T GetUniqueEntityByNamedQuery(String query, Object... params) {
        return CRUDDao.GetUniqueEntityByNamedQuery(query, params);
    }

    @Transactional
    public <T> List<T> GetListByNamedQuery(String query, Object... params) {
        return CRUDDao.GetListByNamedQuery(query, params);
    }

    @Override
    @Transactional(readOnly = true)
    public <T> Long getQueryCount(String query, Object... params) {
        return CRUDDao.getQueryCount(query, params);
    }

    @Override
    @Transactional(readOnly = true)
    public <T> T findByPrimaryKey(Class<T> klass, Serializable id) {
         return CRUDDao.findByPrimaryKey(klass, id);
    }

}

您需要先保存Department在保存之前Employee.

service.save(dept);
service.save(emp);

UPDATE回复您的评论：

为了将员工与部门关联起来，您需要有一个存在的部门。请记住，在您的数据库中，员工对部门有 FK，因此 Hibernate 抱怨的是您试图保存部门不存在的员工，因此您有以下选择：

1. 如果部门是新部门，则必须先保存它，然后再保存员工。
2. 通过查询（例如entityManager.find(id, Department.class)）查找已存储的部门，并在 Employee 对象中使用该对象。
3. Mark as @Cascade您与员工中的部门的关系。