<?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:context="http://www.springframework.org/schema/context"
       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.xsd">
       <context:annotation-config/>
       <context:component-scan base-package="com.jzh"/>

<bean id="dataSource" class="org.apache.commons.dbcp.BasicDataSource" destroy-method="close">
   <property name="driverClassName" value="${jdbc.driverClassName}"/>
   <property name="url" value="${jdbc.url}"/>
   <property name="username" value="${jdbc.username}"/>
   <property name="password" value="${jdbc.password}"/>
 </bean>

<bean id="sessionFactory" class="org.springframework.orm.hibernate3.LocalSessionFactoryBean">
    <property name="dataSource" ref="dataSource"/>
    <property name="mappingResources">
      <list>
        <value>com/jzh/model/User.hbm.xml</value>
      </list>
    </property>
    <property name="hibernateProperties">
      <props>
       <prop key="hibernate.dialect">org.hibernate.dialect.MySQLDialect</prop>
       <prop key="hibernate.show_sql">true</prop>
      </props>
    </property>
  </bean>

</beans>
**********************************************
package com.jzh.service;import javax.annotation.Resource;import org.springframework.stereotype.Component;import com.jzh.dao.UserDAO;
import com.jzh.model.User;
@Component("userManager")
public class UserManager {
private UserDAO userDAO;


public UserDAO getUserDAO() {//此处得不到userDAO return userDAO;
} @Resource(name="userDAO")
public void setUserDAO(UserDAO userDAO) {//但set可以得到
this.userDAO = userDAO;
}
public void saves(User user) 
{
  userDAO.save(user);//此处空指针

}}
******************************************************************
package com.jzh.dao;import javax.annotation.Resource;import org.hibernate.Session;
import org.hibernate.SessionFactory;
import org.hibernate.Transaction;
import org.springframework.stereotype.Component;import com.jzh.model.User;@Component("userDAO")
public class UserDAO {

    private SessionFactory sessionFactory;
public SessionFactory getSessionFactory() {
return sessionFactory;
}
@Resource
public void setSessionFactory(SessionFactory sessionFactory) {
this.sessionFactory = sessionFactory;
}
public void save(User user) 
{
Session session=sessionFactory.getCurrentSession();
Transaction tr=session.beginTransaction();
session.save(user);
tr.commit();
}

}

解决方案 »

  1.   

    成员变量userDao这里标注@Resource(name='userDao'),你标注错了。
      

  2.   

    加到属性上和set方法上是一样的,都报同样的错
      

  3.   


    import java.sql.SQLException;import org.junit.Test;
    import org.springframework.context.support.ClassPathXmlApplicationContext;import com.jzh.model.User;
    import com.jzh.service.UserManager;public class UserManagerTest { @Test
    public void save() throws SQLException
    {
    ClassPathXmlApplicationContext cxa=new ClassPathXmlApplicationContext("beans.xml");
    UserManager um=new UserManager();
    User user=new User();
    user.setAge(20);
    user.setName("10");
    um.saves(user);
    }}
      

  4.   

    就提示在调用userDAO.save(user);这个方法是出现空指针
      

  5.   

    用spring了,你这里UserManager ,User 对象就不能采有new的方式了,要从spring容器中拿对应对象实例才对。
      

  6.   

    怎么从spring容器中拿对应对象实例?  楼主好人 遇到跟你一样的问题了 求方法
      

  7.   


    ApplicationContext ac = new ClassPathXmlApplicationContext("applicationContext.xml");
    Object obj = ac.getBean("beanName");