我是JAVA EE中使用框架的新手
所以当我试图调用实体管理器的方法时
它抛出一个java.lang.NullPointerException
所以我想写我写的东西
这是我的dao实现的一个例子
@Repository
@Transactional
public class IUserDAOImpl implements IUserDAO{
@PersistenceContext
private EntityManager em;
public void addUser(User u) {
em.persist(u);
}
}
这是我的服务
@Service
public class IUserServiceImpl implements IUserService {
@Autowired
private IUserDAO dao = new IUserDAOImpl();
public void setDao(IUserDAO dao) {
this.dao = dao;
}
public void addUser(User u) {
dao.addUser(u);
}
这是我的动作类
public class ConnectionAction extends ActionSupport {
User c = new User("a","b","c",11,"d",true);
@Autowired
public IUserService service;
public String execute() throws Exception {
service.addUser(c);
return SUCCESS;
}
web.xml
<?xml version="1.0" encoding="UTF-8"?>
<web-app id="struts_blank" version="2.4"
xmlns="http://java.sun.com/xml/ns/j2ee"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://java.sun.com/xml/ns/j2ee http://java.sun.com/xml/ns/j2ee/web-app_2_4.xsd">
<display-name>Struts Blank</display-name>
<filter>
<filter-name>struts2</filter-name>
<filter-class>
org.apache.struts2.dispatcher.ng.filter.StrutsPrepareAndExecuteFilter
</filter-class>
</filter>
<filter-mapping>
<filter-name>struts2</filter-name>
<url-pattern>/*</url-pattern>
</filter-mapping>
<context-param>
<param-name>contextConfiguration</param-name>
<param-value>classpath:applicationContext.xml</param-value>
</context-param>
<listener>
<listener-class>
org.springframework.web.context.ContextLoaderListener
</listener-class>
</listener>
</web-app>
applicationContext.xml(虽然它真的搞砸了所以如果你注意到某些错误的plz告诉)
<?xml version="1.0" encoding="UTF-8"?>
<beans xmlns:tx="http://www.springframework.org/schema/tx" xmlns="http://www.springframework.org/schema/beans" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns:context="http://www.springframework.org/schema/context" xmlns:util="http://www.springframework.org/schema/util"
xsi:schemaLocation="
http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-2.5.xsd
http://www.springframework.org/schema/context http://www.springframework.org/schema/context/spring-context-2.5.xsd
http://www.springframework.org/schema/util http://www.springframework.org/schema/util/spring-util-2.0.xsd
http://www.springframework.org/schema/tx http://www.springframework.org/schema/tx/spring-tx-2.5.xsd">
<bean id="dao"
class="com.iticsys.GBO.dao.IUserDAOImpl">
</bean>
<bean id="service"
class="com.iticsys.GBO.services.IUserServiceImpl">
<property name="dao" ref="dao"></property>
</bean>
<bean id="dataSource" class="org.springframework.jdbc.datasource.DriverManagerDataSource">
<property name="driverClassName" value="com.mysql.jdbc.Driver" />
<property name="url" value="jdbc:mysql://localhost:3306/GBO1" />
<property name="username" value="root" />
<property name="password" value="" />
</bean>
<bean id="persistenceUnitManager"
class="org.springframework.orm.jpa.persistenceunit.DefaultPersistenceUnitManager">
<property name="persistenceXmlLocations">
<value>classpath*:META-INF/persistence.xml</value>
</property>
<property name="defaultDataSource" ref="dataSource"> </property>
</bean>
<bean id="entityManagerFactory"
class="org.springframework.orm.jpa.LocalContainerEntityManagerFactoryBean">
<property name="persistenceUnitManager" ref="persistenceUnitManager"></property>
<property name="persistenceXmlLocation" value="classpath*:META-INF/persistence.xml" />
<property name="persistenceUnitName" value="persistenceUnit" />
</bean>
<bean id="transactionManager" class="org.springframework.orm.jpa.JpaTransactionManager">
<property name="entityManagerFactory" ref="entityManagerFactory"></property>
</bean>
<tx:annotation-driven transaction-manager="transactionManager" />
<context:annotation-config></context:annotation-config>
<context:component-scan base-package="com.iticsys.GBO.dao" />
</beans>
的persistence.xml
<persistence xmlns="http://java.sun.com/xml/ns/persistence"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://java.sun.com/xml/ns/persistence http://java.sun.com/xml/ns/persistence/persistence_2_0.xsd"
version="2.0">
<persistence-unit name="persistenceUnit"
transaction-type="RESOURCE_LOCAL">
<provider>org.hibernate.jpa.HibernatePersistenceProvider</provider>
<properties>
<property name="javax.persistence.jdbc.driver" value="com.mysql.jdbc.Driver" />
<property name="javax.persistence.jdbc.url" value="jdbc:mysql://localhost:3306/GBO1" />
<property name="javax.persistence.jdbc.user" value="root" />
<property name="javax.persistence.jdbc.password" value="" />
<property name="hibernate.show_sql" value="true" />
<property name="hibernate.hbm2ddl.auto" value="update" />
</properties>
</persistence-unit>
非常感谢你提前
答案 0 :(得分:0)
在 IUserDAOImpl
类的实现中,您应该实例化EntityManager
的对象
private EntityManager em = new EntityManager(/*constructor parameter if any*/);
或者您可以指定您可以在其他子类中创建的EntityManager
对象的引用,例如IUserServiceImpl
,ConnectionAction
。
为此,您可以在课程 IUserDAOImpl
中添加一个接收该引用的方法。
public void setEntityManager(EntityManager em_ref ){ em = em_ref;}
答案 1 :(得分:0)
public class HibernateUtil {
static Logger logger = Logger.getLogger(HibernateUtil.class);
private static final String FAILED_TO_CREATE_SESSION_FACTORY_OBJECT = "Failed to create sessionFactory object.";
private static final SessionFactory sessionFactory = buildSessionFactory();
private static ServiceRegistry serviceRegistry;
private static SessionFactory buildSessionFactory() {
try {
Configuration configuration = new Configuration();
configuration.configure("./hibernate.cfg.xml");
configuration.addAnnotatedClass(Calzz.class);
serviceRegistry = new StandardServiceRegistryBuilder().applySettings(configuration.getProperties()).build();
return configuration.buildSessionFactory(serviceRegistry);
} catch (Throwable ex) {
logger.debug(FAILED_TO_CREATE_SESSION_FACTORY_OBJECT + ex);
throw new ExceptionInInitializerError(ex);
}
}
public static SessionFactory getSessionFactory() {
return sessionFactory;
}
}