50

这篇文章是JPA 如何在持久化后从数据库中获取值的延续

当我执行以下操作时出现以下异常,我该如何解决?

Not allowed to create transaction on shared EntityManager - use Spring 
transactions or EJB CMT

DAOImpl代码

public void create(Project project) {
        entityManager.persist(project);
        entityManager.getTransaction().commit();
        project = entityManager.find(Project.class, project.getProjectId());
        entityManager.refresh(project);
        System.out.println("Id    -- " + project.getProjectId());
            System.out.println("no -- " + project.getProjectNo());
    }

应用程序上下文.xml

<beans xmlns="http://www.springframework.org/schema/beans"
    xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:tx="http://www.springframework.org/schema/tx"
    xmlns:context="http://www.springframework.org/schema/context"
    xsi:schemaLocation="http://www.springframework.org/schema/beans
http://www.springframework.org/schema/beans/spring-beans-3.0.xsd
http://www.springframework.org/schema/tx
http://www.springframework.org/schema/tx/spring-tx-3.0.xsd
http://www.springframework.org/schema/context
http://www.springframework.org/schema/context/spring-context-3.0.xsd">

    <bean id="DataSource"
        class="org.springframework.jdbc.datasource.DriverManagerDataSource">
        <property name="username" value="scott" />
        <property name="password" value="tiger" />
        <property name="driverClassName" value="oracle.jdbc.driver.OracleDriver" />
        <property name="url" value="jdbc:oracle:thin:@myserver:1521:ORCL" />
    </bean>

    <bean id="entityManagerFactory"
        class="org.springframework.orm.jpa.LocalContainerEntityManagerFactoryBean">
        <property name="dataSource" ref="DataSource" />
        <property name="packagesToScan" value="test.entity" />
        <property name="jpaVendorAdapter">
            <bean class="org.springframework.orm.jpa.vendor.HibernateJpaVendorAdapter">
                <property name="showSql" value="true" />
                <property name="generateDdl" value="false" />
                <property name="databasePlatform" value="org.hibernate.dialect.Oracle10gDialect" />
            </bean>
        </property>
    </bean>

    <context:component-scan base-package="test.net" />

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

     <bean id="transactionManager" class="org.springframework.orm.jpa.JpaTransactionManager">
        <property name="entityManagerFactory" ref="entityManagerFactory" />
    </bean>         

     <context:annotation-config/>

</beans>
4

3 回答 3

56

我想这里的问题是,尽管您已经为事务管理器定义了 bean,但您还没有注释 create() 方法@Transactional来启用 Spring 事务。

还要删除该entityManager.getTransaction().commit();语句,因为现在所有事务管理都将由 Spring 处理,如果您保持该语句不变,那么您将再次收到相同的错误。

于 2013-08-13T06:12:25.167 回答
21

在方法上注入 EntityManagerFactory 而不是 EntityManager 和 javax.transaction.Transactional 注释解决了我的问题,如下所示。

//Autowire EntityManagerFactory
@PersistenceUnit(unitName = "readwrite.config")
private EntityManagerFactory entityManagerFactory;


//Use below code on create/update
EntityManager entityManager = entityManagerFactory.createEntityManager();

entityManager.getTransaction().begin();
if (!ObjectUtils.isEmpty(entity) && !entityManager.contains(entity)) {
   entityManager.persist(entity);
   entityManager.flush();
}
entityManager.getTransaction().commit();
于 2018-11-09T05:19:38.503 回答
5

您需要删除 entityManager.getTransaction().begin() 语句并使用 @Transactional 注释方法,这可以启用弹簧事务

于 2020-02-16T09:27:52.793 回答