java.lang.IllegalStateException：没有可用的事务性 EntityManager

Question

项目使用 Hibernate (JPA)、Spring 和 Maven。我的实体和 DAO 在一个单独的 JAR 中。

pom.xml:

 <project ...>
    ...
    <artifactId>database</artifactId>

    <dependencies>
        <dependency>
            <groupId>org.hibernate</groupId>
            <artifactId>hibernate-entitymanager</artifactId>
            <version>3.5.4-Final</version>
        </dependency>
    </dependencies>
</project>

道：

 public class AbstractDAO<T extends BaseEntity> implements GenericDAO<T> {

    private final Class<T> persistentClass;

    private EntityManager entityManager;

    public AbstractDAO(Class<T> entityClass) {
        super();
        this.persistentClass = entityClass;
    }

    @PersistenceContext
    public void setEntityManager(EntityManager entityManager) {
        this.entityManager = entityManager;
    }

    public EntityManager getEntityManager() {
        return entityManager;
    }

    ...

    public void fooBar() {
       //Exception from this line
       Session session = getEntityManager().unwrap(Session.class);
       ...
    }

    ....

}

我有一个模块，它使用 Spring。

pom.xml:

 <project ...>
...
<artifactId>api</artifactId>

<dependencies>
    <dependency>
        <groupId>org.springframework</groupId>
        <artifactId>spring-core</artifactId>
        <version>${spring.version}</version>
    </dependency>

    <dependency>
        <groupId>org.springframework</groupId>
        <artifactId>spring-context</artifactId>
        <version>${spring.version}</version>
    </dependency>

    <dependency>
        <groupId>org.springframework</groupId>
        <artifactId>spring-orm</artifactId>
        <version>${spring.version}</version>
    </dependency>
    ....
</dependencies>

 ...
</project>

AppContext.xml：

 <bean id="authService" scope="singleton" class="com.test.management.AuthServiceImpl" />

    <bean id="entityManagerFactory" class="org.springframework.orm.jpa.LocalContainerEntityManagerFactoryBean" name="EntityManagerFactory">
        <property name="persistenceUnitName" value="default"></property>
        <property name="dataSource" ref="dataSource"></property>
        <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="${db.dialect}" />
            </bean>
        </property>
    </bean>

    <!-- Values are defined in db.properties -->
    <bean id="dataSource" class="org.apache.commons.dbcp.BasicDataSource" destroy-method="close">
        <property name="driverClassName" value="${db.driver}" />
        <property name="url" value="${db.url}" />
        <property name="username" value="${db.username}" />
        <property name="password" value="${db.password}" />
    </bean>

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

    <tx:annotation-driven />

    <bean id="userDAO" scope="singleton" class="com.test.database.dao.impl.UserDAOImpl">
    </bean>

    <bean class="org.springframework.orm.jpa.support.PersistenceAnnotationBeanPostProcessor" />

</beans>

服务：

 public class AuthServiceImpl implements AuthService {

    @Autowired
    private UserDAO userDAO;

    @Override
    public void authorization() {
        userDAO.fooBar();

    }
}

当我试图从 EntityManager 获取会话时，我发现了这个异常：

 java.lang.IllegalStateException: No transactional EntityManager available
    at org.springframework.orm.jpa.SharedEntityManagerCreator$SharedEntityManagerInvocationHandler.invoke(SharedEntityManagerCreator.java:223)
    at $Proxy121.unwrap(Unknown Source)

原文由 Anthony Tsivarev 发布，翻译遵循 CC BY-SA 4.0 许可协议

Answer 1

您必须使用 @Transactional 注释对该方法进行注释：

 @Transactional
public void fooBar() {
   // Exception from this line
   Session session = getEntityManager().unwrap(Session.class);
   ...
}

并且，如果有必要，例如，当使用普通 Spring 框架时，启用 Spring @Transactional 在 Spring 的 xml 配置文件中使用以下声明进行处理（其中 txManager 是您的经理的 ID）：

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

原文由 poussma 发布，翻译遵循 CC BY-SA 4.0 许可协议

Answer 2

尝试这个 ？

 entityManager=entityManager.getEntityManagerFactory().createEntityManager();
Session session = (Session) entityManager.unwrap(Session.class);

原文由 baybora.oren 发布，翻译遵循 CC BY-SA 3.0 许可协议