问题描述
休眠@version和ManyToOne映射之间有什么关系.
What is the relation between hibernate @version and ManyToOne Mapping.
假设我有两个表Department和Employee.这是百货公司的主表和员工在明细表中.在Employee表中，departmentID被作为外键引用.
Assume that i am having two tables Department and Employee. Here is Deparment is the master tableand Employee in the detail table. In the Employee table, departmentID is reference as foreign key.
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，假设在一页中提取并存储了特定部门在HTTP会话中.
And also, Spring handles the session. So assume that, in one page, particular department is fetched and storedin the HTTP session.
现在在另一页中，我正在尝试执行以下操作
Now in another page, i am trying to do the following
Employee emp = new Employee();
emp.setName('Test')
emp.setDepartment(dept) // already stored in the HTTP session variable
service.save(emp)
现在我遇到以下异常
org.springframework.dao.InvalidDataAccessApiUsageException: object references an unsaved transient instance - save the transient instance before flushing:
然后按如下方式进行更改，就会出现错误
And just it make one change as follow and there is errror
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);
}
}
推荐答案
您需要先保存部门，然后再保存 Employee .
You need to first save the Department before saving the Employee.
service.save(dept);
service.save(emp);
更新以回应您的评论:
要将员工与部门关联，您需要拥有一个部门.请记住，在您的数据库中，该雇员有一个部门的FK，所以Hibernate抱怨的是您正试图用一个不存在的部门来保存一个雇员，因此您有以下选择:
In order to associate an Employee with a Department you need to have a Department that exists. Remember that in your database the Employee has a FK to the Department so what Hibernate is complaining about is that you are trying to save an Employee with a Department that does not exist, so you have these options:
- 如果部门是新部门，则必须先保存它，然后再保存员工.
- 通过诸如EntityManager.find(id，Department.class)之类的查询查找已存储的部门，并在您的Employee对象中使用该对象.
- 在您与Deparment的关系中，将 @Cascade 标记为员工.
- If the Department is a new Department you must save it first before saving the Employee.
- Find an already stored Department through a query such as entityManager.find(id, Department.class) and use that object in your Employee object.
- Mark as @Cascade your relationship with Deparment in the Employee.
这篇关于Hibernate @Version批注的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！