当尝试调用“persist”方法将实体模型保存到 Spring MVC Web 应用程序中的数据库时,出现此错误。在互联网上确实找不到与此特定错误相关的任何帖子或页面。 EntityManagerFactory bean 似乎有问题,但我对 Spring 编程相当陌生,所以对我来说,根据 web.xml 中的各种教程文章,一切似乎都初始化得很好。
调度程序-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>
注册控制器.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";
}
}
我遇到了同样的问题,我将该方法注释为@Transactional
它起作用了。
更新:检查 spring 文档,默认情况下 PersistenceContext 的类型是 Transaction,所以这就是为什么该方法必须是事务性的(http://docs.spring.io/spring/docs/current/spring-framework-reference/html/orm.html):
@PersistenceContext注释有一个可选的属性类型,
默认为 PersistenceContextType.TRANSACTION。这个默认值是
接收共享 EntityManager 代理需要什么。这
替代方案 PersistenceContextType.EXTENDED 是一个完全
不同的事情:这会产生一个所谓的扩展EntityManager,
它不是线程安全的,因此不能同时使用
访问的组件,例如 Spring 管理的单例 bean。扩展
EntityManager 只应该在有状态组件中使用
例如,驻留在会话中,其生命周期为
EntityManager 不依赖于当前事务,而是
完全取决于应用程序。
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