How to inject multiple JPA EntityManager (persistence units) when using Spring(使用Spring时如何注入多个JPA EntityManager(持久化单元))
问题描述
我需要使用一个数据库进行查询(非修改)和一个用于命令(修改).我使用的是 Spring Data JPA,所以我有两个配置类:
I need to use one database for queries (non-modifying) and one for commands (modifying). I am using Spring Data JPA, so I have two configuration classes:
@Configuration
@EnableJpaRepositories(value = "com.company.read",
entityManagerFactoryRef = "readingEntityManagerFactory",
transactionManagerRef = "readingTransactionManager")
@EnableTransactionManagement
public class SpringDataJpaReadingConfiguration {
@Bean(name = "readingEntityManagerFactory")
public EntityManagerFactory readingEntityManagerFactory() {
return Persistence.createEntityManagerFactory("persistence.reading");
}
@Bean(name = "readingExceptionTranslator")
public HibernateExceptionTranslator readingHibernateExceptionTranslator() {
return new HibernateExceptionTranslator();
}
@Bean(name = "readingTransactionManager")
public JpaTransactionManager readingTransactionManager() {
return new JpaTransactionManager();
}
}
@Configuration
@EnableJpaRepositories(value = "com.company.write",
entityManagerFactoryRef = "writingEntityManagerFactory",
transactionManagerRef = "writingTransactionManager")
@EnableTransactionManagement
public class SpringDataJpaWritingConfiguration {
@Bean(name = "writingEntityManagerFactory")
public EntityManagerFactory writingEntityManagerFactory() {
return Persistence.createEntityManagerFactory("persistence.writing");
}
@Bean(name = "writingExceptionTranslator")
public HibernateExceptionTranslator writingHibernateExceptionTranslator() {
return new HibernateExceptionTranslator();
}
@Bean(name = "writingTransactionManager")
public JpaTransactionManager writingTransactionManager() {
return new JpaTransactionManager();
}
}
在我的存储库中,我有时需要使用 EntityManager 来决定这样使用:
In my repository I sometimes need to decide with EntityManager to use like so:
@Repository
public class UserReadingRepository {
@PersistenceContext(unitName = "persistence.reading")
private EntityManager em;
// some useful queries here
}
我正在使用 persistence.xml 中定义的持久性单元名称:
I am using persistence unit's name as defined in my persistence.xml:
<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_2_0.xsd"
version="2.0">
<persistence-unit name="persistence.reading" transaction-type="RESOURCE_LOCAL">
<provider>org.hibernate.jpa.HibernatePersistenceProvider</provider>
<non-jta-data-source>ReadingDS</non-jta-data-source>
<properties>
<property name="hibernate.dialect" value="org.hibernate.dialect.MySQLDialect" />
<property name="hibernate.show_sql" value="true" />
</properties>
</persistence-unit>
<persistence-unit name="persistence.writing" transaction-type="RESOURCE_LOCAL">
<provider>org.hibernate.jpa.HibernatePersistenceProvider</provider>
<non-jta-data-source>WritingDS</non-jta-data-source>
<properties>
<property name="hibernate.dialect" value="org.hibernate.dialect.MySQLDialect" />
<property name="hibernate.show_sql" value="true" />
</properties>
</persistence-unit>
</persistence>
Spring 抛出 org.springframework.beans.factory.NoSuchBeanDefinitionException: No bean named 'persistence.reading' 被定义.奇怪的是,Spring 似乎试图用持久性单元名称实例化 a bean?我是不是配置错了什么?
Spring throws org.springframework.beans.factory.NoSuchBeanDefinitionException: No bean named 'persistence.reading' is defined. Oddly, it looks like Spring tries to instantiate a bean with persistence unit name? Did I misconfigure something?
UPDATE:当我从 @PersistenceContext 注释中删除 unitName = "persistence.reading" 时,我会收到以下错误:org.springframework.beans.factory.NoUniqueBeanDefinitionException:没有定义 [javax.persistence.EntityManagerFactory] 类型的合格 bean:预期的单个匹配 bean 但找到了 2:readingEntityManagerFactory,writingEntityManagerFactory
UPDATE: When I remove unitName = "persistence.reading" from @PersistenceContext annotation, I will get following error instead:
org.springframework.beans.factory.NoUniqueBeanDefinitionException: No qualifying bean of type [javax.persistence.EntityManagerFactory] is defined: expected single matching bean but found 2: readingEntityManagerFactory,writingEntityManagerFactory
更新 2:Rohit 建议(在评论中)改为连接 EntityManagerFactory.所以我尝试执行以下操作:
UPDATE 2: Rohit suggested (in the comment) to wire EntityManagerFactory instead. So I tried to do the following:
@PersistenceUnit(unitName = "persistence.reading")
private EntityManagerFactory emf;
但 Spring 只报告:org.springframework.beans.factory.NoSuchBeanDefinitionException: No bean named 'persistence.reading' is defined
but Spring only reports: org.springframework.beans.factory.NoSuchBeanDefinitionException: No bean named 'persistence.reading' is defined
最终修复:感谢 Vlad 的回答,我能够更新代码以使用以下内容(只需确保您也定义了 dataSource bean):
FINAL FIX:
Thanks to Vlad's answer, I was able to update the code to use the following (just make sure you define your dataSource bean as well):
@Bean(name = "readingEntityManagerFactory")
public EntityManagerFactory readingEntityManagerFactory() {
LocalContainerEntityManagerFactoryBean em = new LocalContainerEntityManagerFactoryBean();
em.setPersistenceUnitName("persistence.reading");
em.setDataSource(dataSource());
em.setPackagesToScan("com.company");
em.setJpaVendorAdapter(new HibernateJpaVendorAdapter());
em.afterPropertiesSet();
return em.getObject();
}
推荐答案
EntityManageFactory 配置不正确.您应该使用 LocalContainerEntityManagerFactoryBean 代替:
The EntityManageFactory is not properly configured. You should use a LocalContainerEntityManagerFactoryBean instead:
@Bean(name = "readingEntityManagerFactory")
public EntityManagerFactory readingEntityManagerFactory() {
LocalContainerEntityManagerFactoryBean em = new LocalContainerEntityManagerFactoryBean();
em.setPersistenceUnitName("persistence.reading");
em.setDataSource(dataSource());
em.setPackagesToScan("com.company");
em.setJpaVendorAdapter(new HibernateJpaVendorAdapter());
em.afterPropertiesSet();
return em.getObject();
}
JpaTransactionManager 也配置错误.它应该是这样的:
Also the JpaTransactionManager is miss-configured too. It should be something like:
@Bean(name = "readingTransactionManager")
public PlatformTransactionManager readingTransactionManager(){
JpaTransactionManager transactionManager = new JpaTransactionManager();
transactionManager.setEntityManagerFactory(readingEntityManagerFactory());
return transactionManager;
}
您需要对 EntityManager 的读取和写入配置执行相同的操作.
You need to do the same for both the reading and the writing EntityManager configurations.
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本文标题为:使用Spring时如何注入多个JPA EntityManager(持久化单元)
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