java中将数据库中数据赋值给Map对象

java中将数据库中数据直接赋值给Map对象，目的是为了方便取值，这种方式需要用到spring框架。 主要代码如下：

1.处理数据是实现类 GetUserDataServiceImp.java

package com.service.Imp;

import java.util.List;
import org.springframework.context.ApplicationContext;
import org.springframework.context.support.ClassPathXmlApplicationContext;
import com.jdbcDao.IjdbcDao;
import com.service.GetUserDataService;
/**
* 处理数据是实现类
* @author Administrator
*/
public class GetUserDataServiceImp implements GetUserDataService{
// IjdbcDao jdbcDao = (IjdbcDao) ApplicationContextUtil.getBean(IjdbcDao.class, "jdbcDao");

ApplicationContext ac=new ClassPathXmlApplicationContext("applicationContext.xml");
IjdbcDao jdbcDao =(IjdbcDao)ac.getBean("jdbcDao");

public List getUserlist(String name) throws Exception {
String sql="select * from t_user t where t.name=?";
List list = jdbcDao.find(sql, new Object[] { name });

return list;
}

}


2.测试类
package com.Test;

import java.util.List;
import java.util.Map;

import com.service.Imp.GetUserDataServiceImp;

/**
* 测试类
* @author Administrator
*/
public class TestClass {
public static void main(String[] args) throws Exception {
GetUserDataServiceImp getUserDataServiceimp=new GetUserDataServiceImp();
List list=getUserDataServiceimp.getUserlist("小明");
String StrName="";//姓名
String StrAge="";//年龄
if(list!=null&&list.size()>0){
Map result =null;
for(int i=0;i<list.size();i++){
result = (Map) list.get(i);//将结果赋予Map
StrName=(String) result.get("name");
StrAge=(String) result.get("age");
System.out.println("返回结果：name===="+StrName+" ;age====="+StrAge);
}
}
}

}

详细源代码：http://download.csdn.net/detail/u012255097/9744965