我想要做的是通过XML注入几乎与通过A @PersistenceContext注释完成的方式相同.我需要这个,因为我需要将不同的实体管理器注入到同一个DAO中.数据库彼此镜像,我宁愿有1个基类,对于那个基类的实例,然后创建多个类,这样我就可以使用@PersistenceContext注释.
这是我的例子.这就是我现在正在做的事情并且它有效.
public class ItemDaoImpl {
protected EntityManager entityManager;
public List<Item> getItems() {
Query query = entityManager.createQuery("select i from Item i");
List<Item> s = (List<Item>)query.getResultList();
return s;
}
public void setEntityManger(EntityManager entityManager) {
this.entityManager = entityManager;
}
}
@Repository(value = "itemDaoStore2")
public class ItemDaoImplStore2 extends ItemDaoImpl {
@PersistenceContext(unitName = "persistence_unit_2")
public void setEntityManger(EntityManager entityManager) {
this.entityManager = entityManager;
}
}
@Repository(value = "itemDaoStore1")
public class ItemDaoImplStore1 extends ItemDaoImpl {
@PersistenceContext(unitName = "persistence_unit_1")
public void setEntityManger(EntityManager entityManager) {
this.entityManager = …当一个Entity映射到另一个在其子类上有直接实现的Entity时,我遇到了问题.请参阅以下示例映射:
@Entity
class Location {
@OneToOne
@JoinColumn(...)
private Person person;
}
@Entity
@Inheritance(strategy=InheritanceType.SINGLE_TABLE)
@DiscriminatorColumn(name="person_type",discriminatorType=DiscriminatorType.STRING)
abstract class Person {
}
@Entity
@DiscriminatorValue("M")
class Man extends Person {
...
}
@Entity
@DiscriminatorValue("W")
class Woman extends Person {
...
}
现在,这就是我在数据库表上的内容:
location-table:id = 1,person_id = 1 person-table:id = 1,person_type ="M"
当我使用实体管理器检索位置时,hibernate会抛出一个异常,说我无法实例化抽象类或接口.
Location location = entityManager.find(Location.class, 1L);
Hibernate抛出此错误:
javax.persistence.PersistenceException: org.hibernate.InstantiationException: Cannot instantiate abstract class or interface: Person
at org.hibernate.ejb.AbstractEntityManagerImpl.throwPersistenceException(AbstractEntityManagerImpl.java:630)
at org.hibernate.ejb.AbstractEntityManagerImpl.find(AbstractEntityManagerImpl.java:195)
at ......
at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method)
at sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:39)
at sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:25)
at java.lang.reflect.Method.invoke(Method.java:585)
at org.junit.internal.runners.TestMethod.invoke(TestMethod.java:59) …根据其JavaDoc,PersistenceAnnotationBeanPostProcessor似乎负责使用注释@PersistenceContext注入EntityManager.它似乎暗示如果没有在Spring应用程序上下文xml中声明这个bean,@ PerersContext注释将不起作用.
但是,根据我的实验,这不是事实.
<persistence xmlns="http://java.sun.com/xml/ns/persistence"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://java.sun.com/xml/ns/persistence
http://java.sun.com/xml/ns/persistence/persistence_1_0.xsd"
version="1.0">
<persistence-unit name="default" transaction-type="RESOURCE_LOCAL" />
</persistence>
<context:component-scan base-package="com.test.dao" />
<bean id="entityManagerFactory" class="org.springframework.orm.jpa.LocalContainerEntityManagerFactoryBean">
<property name="dataSource" ref="dataSource"/>
<property name="persistenceUnitName" value="default"/>
<property name="jpaVendorAdapter">
<bean class="org.springframework.orm.jpa.vendor.HibernateJpaVendorAdapter">
<property name="showSql" value="true"/>
<property name="generateDdl" value="true"/>
<property name="databasePlatform" value="org.hibernate.dialect.DerbyDialect"/>
</bean>
</property>
</bean>
<bean id="dataSource" class="org.springframework.jdbc.datasource.DriverManagerDataSource">
<property name="driverClassName" value="org.apache.derby.jdbc.ClientDriver"/>
<property name="url" value="jdbc:derby://localhost:1527/c:\derbydb\mydb"/>
<property name="username" value="APP"/>
<property name="password" value="APP"/>
</bean>
<tx:annotation-driven/>
<bean id="transactionManager" class="org.springframework.orm.jpa.JpaTransactionManager">
<property name="entityManagerFactory" ref="entityManagerFactory" />
</bean>
<!--
<bean id="persistenceAnnotation" class="org.springframework.orm.jpa.support.PersistenceAnnotationBeanPostProcessor" />
-->
@Repository("userDao")
public class UserDaoImpl …我有两个Entitymanagerbean配置.每个指向具有不同模式的单独数据库(一个是Oracle,另一个是内存中的H2)
我该怎么做才能解决Entitymanager应该用于每个存储库的模糊性？现在我收到这个错误:
No unique bean of type [javax.persistence.EntityManagerFactory] is defined:
expected single bean but found 2
我想我可以通过使用类似的东西来提供快速修复
<jpa:repositories base-package="com.foo.repos.ora"
entity-manager-factory-ref="entityManagerFactoryA">
<jpa:repositories base-package="com.foo.repos.m2"
entity-manager-factory-ref="entityManagerFactoryB">
但希望有更好的解决方案.
编辑:
我给你一个当前场景的想法:
Spring-Config:有两个EM
<jpa:repositories base-package="com.foo.repos.ora" entity-manager-factory-ref="entityManagerFactory"/>
<jpa:repositories base-package="com.foo.repos.m2" entity-manager-factory-ref="entityManagerFactory2"/>
<context:component-scan base-package="com.foo" /> ....
这里的所有内容都在"package com.foo.repos.ora"中.遵循如何创建自定义存储库的模式我得到两个接口'ARepository','ARepositoryCustom'及其实现'ARepositoryImpl'就像这样
@Repository
public interface ARepository extends ARepositoryCustom, JpaRepository<myEntity, BigDecimal>, QueryDslPredicateExecutor {
}
public interface ARepositoryCustom {
FooBar lookupFooBar()
}
public class ARepositoryImpl extends QueryDslRepositorySupport implements ARepositoryCustom {
ARepositoryImpl(Class<?> domainClass) {
super(domainClass.class)
}
ARepositoryImpl() {
this(myEntity.class)
}
@Override
FooBar …在我的具体案例中,我正在使用一个鉴别器列策略.这意味着我的JPA实现(Hibernate)创建了一个带有特殊DTYPE列的用户表.此列包含实体的类名.例如,我的users表可以包含TrialUser和PayingUser的子类.这些类名称将位于DTYPE列中,以便当EntityManager从数据库加载实体时,它知道要实例化的类的类型.
我尝试过两种转换实体类型的方法,两种方式都像脏黑客一样:
#1的问题是当您手动更改此列时,JPA不知道如何刷新/重新附加此实体到持久性上下文.它需要一个TrialUser ID为1234,而不是一个PayingUser ID为1234它失败了.在这里,我可以做一个EntityManager.clear()并分离所有实体/清除Per.上下文,但由于这是一个Service bean,它将擦除系统所有用户的挂起更改.
#2的问题在于,当您删除TrialUser时,您设置为Cascade = ALL的所有属性也将被删除.这很糟糕,因为您只是尝试交换其他用户,而不是删除所有扩展对象图.
更新1:#2的问题使我几乎无法使用,所以我放弃了尝试让它工作.更优雅的黑客肯定是#1,我在这方面取得了一些进展.关键是首先获得对底层Hibernate Session的引用(如果您使用Hibernate作为JPA实现)并调用Session.evict(user)方法从持久性上下文中仅删除该单个对象.不幸的是,没有纯粹的JPA支持.以下是一些示例代码:
// Make sure we save any pending changes
user = saveUser(user);
// Remove the User instance from the persistence context
final Session session = (Session) entityManager.getDelegate();
session.evict(user);
// Update the DTYPE
final String sqlString = "update user set user.DTYPE = '" + targetClass.getSimpleName() + "' where user.id = …因为调用flush()来使每个实体从内存持久存储到数据库.因此,如果我使用调用太多不必要的flush(),则可能需要很长时间,因此不是性能的好选择.这是一个我不知道何时调用flush()的场景？
//Order and Item have Bidirectional Relationships
Order ord = New ord("my first order");
Item item = New Item("tv",10);
//...process item and ord object
em.persist(ord);//em is an instance of EntityManager
em.flush();// No.1 flush()
item.setOrder(ord);
em.persist(item);
Set<Item> items= new HashSet<Item>();
items.add(item);
ord.setItems(items);
em.flush();// No.2 flush()
我的问题是:拨打第1号同花顺是否可以避免？
我担心的是:为了执行item.setOrder(ord),我们需要一个ord的数据库ID.并且只调用em.persist(ord)无法生成数据库ID,因此我必须在item.setOrder(ord)之前调用em.flush().那么你们的观点是什么？
提前致谢.
与我的另一个问题有点相关应该从数据访问层或接口返回原始的Hibernate注释POJO吗？,我在创建很好的分离层方面很有经验,但没有使用Hibernate或J2EE/JPA.我一直在查看文档和教程,并对如何以优雅的方式使用EntityManger感到困惑,因为它似乎负责两个事务(我想在我的服务层执行)和持久性方法(我想要的)保持在数据访问层).我应该在服务层创建它并将其注入数据访问层,还是有更好的方法？下面的伪java大致显示了我正在考虑做的事情.
编辑:我的下面的伪代码主要取自hibernate JPA教程,并针对图层分离进行了修改,并未反映出该产品是在EJB容器(Glassfish)中运行的.在您的答案中,请提供在Glassfish中运行的代码或同等代码的最佳实践和代码示例.
MyService
{
setup()
{
EntityManagerFactory entityManagerFactory = Persistence.createEntityManagerFactory( "Something" ); //is the String you pass in important?
entityManager = entityManagerFactory.createEntityManager();
}
myServiceMethod()
{
entityManager.getTransaction().begin();
MyDao.setEntityManager(entityManagerFactory);
MyDao.doSomething();
MyDao.doSomethingElse();
entityManager.getTransaction().commit();
entityManager.close();
}
}
MyDao
{
doSomething()
{
entityManager.persist(...); //etc
}
}
我有2个实体:Account和AccountRole.
public class Account {
private AccountRole accountRole;
@ManyToOne(cascade = CascadeType.PERSIST, fetch = FetchType.EAGER)
public AccountRole getAccountRole() {
return accountRole;
}
.
public class AccountRole {
private Collection<Account> accounts = new ArrayList<Account>();
@OneToMany(mappedBy = "accountRole", fetch = FetchType.EAGER)
public Collection<Account> getAccounts() {
return accounts;
}
当我从数据库中获取accountRole并试图坚持我的问题时出现问题Account.此时我刚创建了我的帐户,角色已经存在于db中.
AccountRole role = accountService.getRoleFromDatabase(AccountRoles.ROLE_USER);
account.setAccountRole(role);
//setting both ways, as suggested
public void setAccountRole(AccountRole accountRole) {
accountRole.addAccount(this);
this.accountRole = accountRole;
}
entityManager.persist(account); // finally in my DAO
我读到这个: …
我将Wildfly 8.1与包含实体的EJB项目(EJB 3.2)一起使用.当我尝试将实体管理器注入我的一个Beans时,我得到以下内容:
JBAS011440: Can't find a persistence unit named null in deployment \"EntitiesProject.jar\""},
"JBAS014771: Services with missing/unavailable dependencies" => [
"jboss.deployment.unit.\"EntitiesProject.jar\".weld.weldClassIntrospector is missing [jboss.deployment.unit.\"EntitiesProject.jar\".beanmanager]",
"jboss.naming.context.java.comp.EntitiesProject.EntitiesProject.MitarbeiterDAO.HandleDelegate is missing [jboss.naming.context.java.comp.EntitiesProject.EntitiesProject.MitarbeiterDAO]",
"jboss.naming.context.java.comp.EntitiesProject.EntitiesProject.MitarbeiterDAO.InstanceName is missing [jboss.naming.context.java.comp.EntitiesProject.EntitiesProject.MitarbeiterDAO]",
"jboss.naming.context.java.comp.EntitiesProject.EntitiesProject.MitarbeiterDAO.InAppClientContainer is missing [jboss.naming.context.java.comp.EntitiesProject.EntitiesProject.MitarbeiterDAO]",
"jboss.naming.context.java.comp.EntitiesProject.EntitiesProject.MitarbeiterDAO.Validator is missing [jboss.naming.context.java.comp.EntitiesProject.EntitiesProject.MitarbeiterDAO]",
"jboss.naming.context.java.comp.EntitiesProject.EntitiesProject.MitarbeiterDAO.ORB is missing [jboss.naming.context.java.comp.EntitiesProject.EntitiesProject.MitarbeiterDAO]",
"jboss.naming.context.java.comp.EntitiesProject.EntitiesProject.MitarbeiterDAO.ValidatorFactory is missing [jboss.naming.context.java.comp.EntitiesProject.EntitiesProject.MitarbeiterDAO]"
]
这是我的SessionBean,我尝试注入EntityManager:
@Stateless
public class MitarbeiterDAO implements MitarbeiterDAORemote {
@PersistenceContext
private EntityManager em;
public int writeMitarbeiterToDB(Mitarbeiter m)
{
em.persist(m);
return m.getId();
}
}
我已经persistence.xml为"project-name"/ ejbModule/META-INF 指定了以下文件.
<persistence-unit name="mitarbeiter">
<provider>org.hibernate.ejb.HibernatePersistence</provider> …我已经配置了两个到数据库的连接.一个连接称为用户,另一个称为客户端.这是config.yml文件中的配置:
doctrine:
dbal:
default_connection: client
connections:
client:
driver: pdo_mysql
host: '%client_database_host%'
port: '%client_database_port%'
dbname: '%client_database_name%'
user: '%client_database_user%'
password: '%client_database_password%'
charset: UTF8
mapping_types:
enum: string
user:
driver: pdo_mysql
host: '%user_database_host%'
port: '%user_database_port%'
dbname: '%user_database_name%'
user: '%user_database_user%'
password: '%user_database_password%'
charset: UTF8
mapping_types:
enum: string
orm:
auto_generate_proxy_classes: '%kernel.debug%'
default_entity_manager: ~
entity_managers:
client:
naming_strategy: doctrine.orm.naming_strategy.underscore
mappings:
ProjectModelBundle: ~
connection: client
user:
naming_strategy: doctrine.orm.naming_strategy.underscore
mappings:
BaseModelBundle: ~
ProjectModelBundle: ~
connection: user
但无论如何,我总是得到第一个实体经理.这就是我在服务中使用实体管理器的方式:
# BASE
htec.project_model_bundle.repository.database.client_base:
class: Project\BaseModelBundle\Repository\Database\DatabaseRepository
arguments: ['@service_container', '@doctrine.orm.client_entity_manager', '@form.factory']
htec.project_model_bundle.repository.database.user_base:
class: Project\BaseModelBundle\Repository\Database\DatabaseRepository
arguments: …entitymanager ×10
hibernate ×5
java ×5
jpa ×5
spring ×2
annotations ×1
doctrine ×1
flush ×1
java-ee ×1
one-to-one ×1
orm ×1
spring-data ×1
symfony ×1
symfony-3.3 ×1
transactions ×1
wildfly ×1
xml ×1