저는 Spring에서 새롭기 때문에 Spring 3.0M3을 Mac의 Tomcat 서버에두고 Eclipse 프로젝트를 작성하고 Hello World를 작성한 다음 객체는 Hibernate와 지속된다. 내 MySQL 서버에서 User 테이블을 만들었고, 모든 getter와 setter를 가진 User 객체를 만들었다. (자바가 Objective-C에서 큐를 가져 와서 동적 속성을 추가하기를 바란다.) UserDao 객체를 검색하여 User를 저장하고 Bean을 구성합니다. 세션이 초기화되지 않는다는 것을 제외하고는 꽤 잘됩니다. 왜 이런거야? 이 컴퓨터에서이 자격 증명을 사용하여 데이터베이스에 액세스 할 수 있습니다.Hibernate/Spring3 : 프록시를 초기화 할 수 없음 - no Session
나는 이것들이 아마 정상적인 초보자 물건이라는 것을 알았지 만, 인터넷 검색이 최대 절전 모드 2에서 3으로 전환 할 때 중간에 세션을 잃어 버리고있는 사람들이었다고 나는 오류에서 발견했다. 나는 전환하지 않는다. , 그리고 내가 말할 수있는 한 세션은 결코 만들어지지 않는다. 여기
<bean id="dataSource" class="org.apache.commons.dbcp.BasicDataSource">
<property name="driverClassName" value="com.mysql.jdbc.Driver" />
<property name="url" value="jdbc:mysql://10.0.0.3:3306/HelloDB" />
<property name="username" value="hello" />
<property name="password" value="w0rld" />
<property name="initialSize" value="2" />
<property name="maxActive" value="5" />
</bean>
<bean id="sessionFactory" class="org.springframework.orm.hibernate3.annotation.AnnotationSessionFactoryBean">
<property name="dataSource" ref="dataSource"/>
<property name="annotatedClasses">
<list>
<value>com.saers.data.entities.User</value>
</list>
</property>
<property name="hibernateProperties">
<props>
<prop key="hibernate.dialect">org.hibernate.dialect.MySQLDialect</prop>
<prop key="hibernate.show_sql">true</prop>
<prop key="hibernate.lazy">false</prop>
</props>
</property>
</bean>
<bean id="transactionManager" class="org.springframework.orm.hibernate3.HibernateTransactionManager">
<property name="sessionFactory" ref="sessionFactory"/>
</bean>
<bean id="userDAO" class="com.saers.data.dao.UserDao">
<property name="sessionFactory" ref="sessionFactory"/>
</bean>
<bean id="txProxyTemplate" class="org.springframework.transaction.interceptor.TransactionProxyFactoryBean" abstract="true">
<property name="transactionManager" ref="transactionManager"/>
<property name="transactionAttributes">
<props>
<prop key="add*">PROPAGATION_REQUIRED</prop>
<prop key="update*">PROPAGATION_REQUIRED</prop>
<prop key="delete*">PROPAGATION_REQUIRED</prop>
<prop key="*">PROPAGATION_SUPPORTS,readOnly</prop>
</props>
</property>
</bean>
<bean id="userService" parent="txProxyTemplate">
<property name="target">
<bean class="com.saers.business.UserServiceImpl">
<property name="userDao" ref="userDAO"/>
</bean>
</property>
<property name="proxyInterfaces" value="com.saers.business.UserService"/>
</bean>
<bean name="/" class="com.saers.view.web.controller.HelloWorldController">
<property name="userService" ref="userService"/>
</bean>
내 사용자 클래스 :
package com.saers.data.entities;
import java.util.Date;
import java.io.Serializable;
import javax.persistence.*;
@Entity
@Table(name = "User")
public class User implements Serializable {
private static final long serialVersionUID = -6123654414341191669L;
@Id
@Column(name = "WebUserId")
private String WebUserId;
@Column(name = "Name")
private String Name;
/**
* @return the webUserId
*/
public synchronized String getWebUserId() {
return WebUserId;
}
/**
* @param webUserId the webUserId to set
*/
public synchronized void setWebUserId(String webUserId) {
WebUserId = webUserId;
}
/**
* @return the name
*/
public synchronized String getName() {
return Name;
}
/**
* @param name the name to set
*/
public synchronized void setName(String name) {
Name = name;
}
}
그리고 여기 내 UserDao입니다 여기에
SEVERE: Servlet.service() for servlet HelloApp threw exception
org.hibernate.LazyInitializationException: could not initialize proxy - no Session
at org.hibernate.proxy.AbstractLazyInitializer.initialize(AbstractLazyInitializer.java:57)
at org.hibernate.proxy.AbstractLazyInitializer.getImplementation(AbstractLazyInitializer.java:111)
at org.hibernate.proxy.pojo.cglib.CGLIBLazyInitializer.invoke(CGLIBLazyInitializer.java:150)
at com.saers.data.entities.User$$EnhancerByCGLIB$$c2f16afd.getName(<generated>)
at com.saers.view.web.controller.HelloWorldController.handleRequestInternal(HelloWorldController.java:22)
at org.springframework.web.servlet.mvc.AbstractController.handleRequest(AbstractController.java:153)
at org.springframework.web.servlet.mvc.SimpleControllerHandlerAdapter.handle(SimpleControllerHandlerAdapter.java:48)
at org.springframework.web.servlet.DispatcherServlet.doDispatch(DispatcherServlet.java:763)
at org.springframework.web.servlet.DispatcherServlet.doService(DispatcherServlet.java:709)
at org.springframework.web.servlet.FrameworkServlet.processRequest(FrameworkServlet.java:613)
at org.springframework.web.servlet.FrameworkServlet.doGet(FrameworkServlet.java:525)
at javax.servlet.http.HttpServlet.service(HttpServlet.java:617)
at javax.servlet.http.HttpServlet.service(HttpServlet.java:717)
at org.apache.catalina.core.ApplicationFilterChain.internalDoFilter(ApplicationFilterChain.java:290)
at org.apache.catalina.core.ApplicationFilterChain.doFilter(ApplicationFilterChain.java:206)
at org.apache.catalina.core.StandardWrapperValve.invoke(StandardWrapperValve.java:233)
at org.apache.catalina.core.StandardContextValve.invoke(StandardContextValve.java:191)
at org.apache.catalina.core.StandardHostValve.invoke(StandardHostValve.java:128)
at org.apache.catalina.valves.ErrorReportValve.invoke(ErrorReportValve.java:102)
at org.apache.catalina.core.StandardEngineValve.invoke(StandardEngineValve.java:109)
at org.apache.catalina.connector.CoyoteAdapter.service(CoyoteAdapter.java:293)
at org.apache.coyote.http11.Http11Processor.process(Http11Processor.java:849)
at org.apache.coyote.http11.Http11Protocol$Http11ConnectionHandler.process(Http11Protocol.java:583)
at org.apache.tomcat.util.net.JIoEndpoint$Worker.run(JIoEndpoint.java:454)
at java.lang.Thread.run(Thread.java:637)
는 관련 콩에 대한 내 서블릿 설정이다 : 여기
내 오류입니다 :
package com.saers.data.dao;
import java.util.List;
import com.saers.data.entities.User;
import org.springframework.orm.hibernate3.support.*;
public class UserDao extends HibernateDaoSupport {
public void saveUser(User user) {
getHibernateTemplate().saveOrUpdate(user);
}
public User lookupUser(String WebUserId) {
User user = getHibernateTemplate().load(User.class, WebUserId);
return user;
return user;
}
}
public class UserServiceImpl implements UserService {
private UserDao userDao;
public void setUserDao(UserDao userDao) {
this.userDao = userDao;
}
@Override
public User lookupUser(String webUserId) {
return userDao.lookupUser(webUserId);
}
}
그리고 마지막으로, 여기 내 HelloWorldController입니다 :
package com.saers.view.web.controller;
import com.saers.data.entities.User;
import com.saers.business.UserService;
import org.springframework.web.servlet.mvc.*;
import org.springframework.web.servlet.*;
import javax.servlet.http.*;
import org.apache.commons.logging.*;
public class HelloWorldController extends AbstractController {
protected final Log logger = LogFactory.getLog(getClass());
@Override
public ModelAndView handleRequestInternal(HttpServletRequest request, HttpServletResponse response) throws Exception {
logger.info("Get bean");
User user = userService.lookupUser("helloUser");
logger.info("Found out that this user was last changed " + user.getName());
logger.info("Return View");
ModelAndView mv = new ModelAndView("HelloWorld.jsp", "user", user);
return mv;
}
private UserService userService = null;
public void setUserService(UserService userService) {
this.userService = userService;
}
}
난 당신이 내가 사용할 수있는 몇 가지 좋은 조언을 희망
여기 내 UserService 인터페이스
public interface UserService {
public User lookupUser(String webUserId);
public void setUserDao(UserDao userDao);
}
과 그 구현의 : -) 코드에서 코드가 잘못되었거나 스프링이 아닌 것으로 느껴지면 t에 대해 듣고 싶습니다. 모자도.
건배
Nik
업데이트 된 답변보기 – skaffman