我得到这个错误时,试图调用“持久”方法来保存实体模型到数据库在我的Spring MVC web应用程序。在互联网上找不到任何与这个特定错误有关的帖子或页面。似乎EntityManagerFactory bean有问题,但我对Spring编程相当陌生,所以对我来说,似乎一切都初始化得很好,根据各种教程文章在web。

dispatcher-servlet.xml

<?xml version="1.0" encoding="UTF-8"?>
<beans xmlns="http://www.springframework.org/schema/beans"
 xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:mvc="http://www.springframework.org/schema/mvc"
 xmlns:context="http://www.springframework.org/schema/context"
 xmlns:jpa="http://www.springframework.org/schema/data/jpa"
xsi:schemaLocation="
 http://www.springframework.org/schema/mvc 
 http://www.springframework.org/schema/mvc/spring-mvc-4.0.xsd
 http://www.springframework.org/schema/beans 
 http://www.springframework.org/schema/beans/spring-beans-4.0.xsd
 http://www.springframework.org/schema/context 
  http://www.springframework.org/schema/context/spring-context-4.0.xsd
  http://www.springframework.org/schema/jdbc
  http://www.springframework.org/schema/jdbc/spring-jdbc-3.2.xsd
  http://www.springframework.org/schema/data/jpa
  http://www.springframework.org/schema/data/jpa/spring-jpa-1.3.xsd
  http://www.springframework.org/schema/data/repository
  http://www.springframework.org/schema/data/repository/spring-repository-1.5.xsd
  http://www.springframework.org/schema/jee
  http://www.springframework.org/schema/jee/spring-jee-3.2.xsd">

    <context:component-scan base-package="wymysl.Controllers" />
    <jpa:repositories base-package="wymysl.repositories"/> 
    <context:component-scan base-package="wymysl.beans" /> 
    <context:component-scan base-package="wymysl.Validators" /> 
    <bean
     class="org.springframework.orm.jpa.support.PersistenceAnnotationBeanPostProcessor" />
     <bean class="org.springframework.orm.hibernate4.HibernateExceptionTranslator"/>

     <bean id="passwordValidator" class="wymysl.Validators.PasswordValidator"></bean>

     <bean id="dataSource"
        class="org.springframework.jdbc.datasource.DriverManagerDataSource">

        <property name="driverClassName" value="oracle.jdbc.driver.OracleDriver" />
        <property name="url" value="jdbc:oracle:thin:@localhost:1521:xe" />
        <property name="username" value="system" />
        <property name="password" value="polskabieda1" />
    </bean>

 <bean id="entityManagerFactory" class="org.springframework.orm.jpa.LocalContainerEntityManagerFactoryBean">
    <property name="persistenceXmlLocation" value="classpath:./META-INF/persistence.xml" />
    <property name="dataSource" ref="dataSource" />

    <property name="jpaVendorAdapter">
        <bean class="org.springframework.orm.jpa.vendor.HibernateJpaVendorAdapter">
            <property name="databasePlatform" value="org.hibernate.dialect.H2Dialect" />
            <property name="showSql" value="true" />
            <property name="generateDdl" value="false" />
        </bean>
    </property>
    <property name="jpaProperties">
        <props>
            <prop key="hibernate.max_fetch_depth">3</prop>
            <prop key="hibernate.jdbc.fetch_size">50</prop>
            <prop key="hibernate.jdbc.batch_size">10</prop>
        </props>
    </property>
</bean>

    <mvc:annotation-driven />

    <bean id="messageSource" class="org.springframework.context.support.ReloadableResourceBundleMessageSource">
    <property name="basename" value="classpath:messages" />
</bean>

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


    <bean class="org.springframework.web.servlet.view.InternalResourceViewResolver">
    <property name="prefix">
        <value>/WEB-INF/jsp/</value>
    </property>
    <property name="suffix">
        <value>.jsp</value>
    </property>
</bean>

    <mvc:resources mapping="/resources/**" location="/resources/" />
    <mvc:resources mapping="/resources/*" location="/resources/css/"  
    cache-period="31556926"/>



</beans>

RegisterController.java

@Controller
public class RegisterController {

    @PersistenceContext
    EntityManager entityManager;

    @Autowired
    PasswordValidator passwordValidator;

    @InitBinder
    private void initBinder(WebDataBinder binder) {
        binder.setValidator(passwordValidator);
    }

    @RequestMapping(value = "/addUser", method = RequestMethod.GET)
    public String register(Person person) {


        return "register";

    }

    @RequestMapping(value = "/addUser", method = RequestMethod.POST)
    public String register(@ModelAttribute("person") @Valid @Validated Person person, BindingResult result) {
        if(result.hasErrors()) {
            return "register";
        } else {
            entityManager.persist(person);
            return "index";

        }




    }

当前回答

如果你有

@Transactional // Spring Transactional
class MyDao extends Dao {
}

和父类

class Dao {
    public void save(Entity entity) { getEntityManager().merge(entity); }
}

你打电话给我

@Autowired MyDao myDao;
myDao.save(entity);

你不会得到一个Spring TransactionInterceptor(给你一个事务)。

这是你需要做的:

@Transactional 
class MyDao extends Dao {
    public void save(Entity entity) { super.save(entity); }
}

难以置信,但事实如此。

其他回答

我有这个问题好几天了,我在网上找不到任何帮助我的东西,我把我的答案贴在这里,以防它能帮助到其他人。

在我的例子中,我正在处理一个通过远程调用的微服务,并且远程代理没有拾取服务级别上的@Transactional注释。

在服务层和dao层之间添加一个委托类,并将委托方法标记为事务性,这为我解决了这个问题。

在类级别为测试类添加org.springframework.transaction.annotation.Transactional注释为我解决了这个问题。

我也有同样的问题,我在applicationContext.xml中添加了tx:annotation-driven,它工作了。

我移除了模式

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

要做到这一点

当我在错误的方法/动作级别上使用@Transaction时,我有相同的错误代码。

methodWithANumberOfDatabaseActions() { 
   methodA( ...)
   methodA( ...)
}

@Transactional
void methodA( ...) {
  ... ERROR message
}

当然,我必须将@Transactional放在methodWithANumberOfDatabaseActions()方法的上方。

在我的例子中,这解决了错误消息。