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2024-10-11 07:00:03

Spring -当前线程没有实际事务可用的EntityManager -不能可靠地处理'persist'调用

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

这对我们有帮助，也许将来也能帮助到其他人。@Transaction对我们不起作用，但是这个起作用了:

@ConditionalOnMissingClass（“org.springframework.orm.jpa.JpaTransactionManager”）

2019-09-10 13:45:57

其他回答

当我从Junit测试用例中执行Spring JPA deleteAll()方法时，我得到了相同的错误。我简单地使用了deleteInBatch() & deleteAllInBatch()，它完美地工作。我们不需要在测试用例级别标记@Transactional。

2020-12-09 05:12:26

boardRepo.deleteByBoardId (id);

面对同样的问题。transactionrequiredexception:当前线程没有实际事务可用的EntityManager

我通过在控制器/服务上面添加@Transactional注释来解决这个问题。

2018-11-30 16:35:09

当我从同一个组件中的非事务性方法访问一个已经带有事务注释的方法时，我也犯了同样的错误:

Before:
    @Component
    public class MarketObserver {
        @PersistenceContext(unitName = "maindb")
        private EntityManager em;

        @Transactional(value = "txMain", propagation = Propagation.REQUIRES_NEW)
        public void executeQuery() {
          em.persist(....);
        }


        @Async
        public void startObserving() {
          executeQuery(); //<-- Wrong
        }
    }

    //In another bean:
     marketObserver.startObserving();

我通过在自引用组件上调用executeQuery()来修复这个错误:

Fixed version:
    @Component
    public class MarketObserver {
        @PersistenceContext(unitName = "maindb")
        private EntityManager em;

        @Autowired
        private GenericApplicationContext context;

        @Transactional(value = "txMain", propagation = Propagation.REQUIRES_NEW)
        public void executeQuery() {
          em.persist(....);
        }


        @Async
        public void startObserving() {
          context.getBean(MarketObserver.class).executeQuery(); //<-- Works
        }
    }
2017-12-20 18:40:39

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。

2020-05-12 13:43:07

对我们来说，问题在于多个配置文件中的相同上下文设置。检查您没有在多个配置文件中复制以下内容。

<context:property-placeholder location="classpath*:/module.properties"/>
<context:component-scan base-package="...." />
2017-04-19 12:41:18

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