jsp Hibernate入门教程第1/3页

答案：例如： 
复制代码 代码如下:

HibernateTest.java 
import onlyfun.caterpillar.*; 
import net.sf.hibernate.*; 
import net.sf.hibernate.cfg.*; 
import java.util.*; 
public class HibernateTest { 
public static void main(String[] args) throws HibernateException { 
SessionFactory sessionFactory = new Configuration().configure().buildSessionFactory(); 
Session session = sessionFactory.openSession(); 
List users = session.find("from User"); 
User updated = null; 
for (ListIterator iterator = users.listIterator(); iterator.hasNext(); ) { 
User user = (User) iterator.next(); 
if(updated == null) 
updated = user; 
System.out.println(user.getName() + 
"\n\tAge: " + user.getAge() + 
"\n\tSex: " + user.getSex()); 
} 
updated.setName("justin"); 
session.flush(); 
users = session.find("from User"); 
session.close(); 
sessionFactory.close(); 
for (ListIterator iterator = users.listIterator(); iterator.hasNext(); ) { 
User user = (User) iterator.next(); 
System.out.println(user.getName() + 
"\n\tAge: " + user.getAge() + 
"\n\tSex: " + user.getSex()); 
} 
} 
} 

　这个程序会显示数据表中的所有数据，并将数据表中的第一笔数据更新，一个执行的结果如下： 
log4j:WARN No appenders could be found for logger (net.sf.hibernate.cfg.Environment). 
log4j:WARN Please initialize the log4j system properly. 
Hibernate: select user0_.user_id as user_id, user0_.name as name, user0_.易做图 as 易做图, user0_.age as age from USER user0_ 
caterpillar 
Age: 28 
Sex: M 
momor 
Age: 25 
Sex: F 
Bush 
Age: 25 
Sex: M 
Becky 
Age: 35 
Sex: F 
Hibernate: update USER set name=?, 易做图=?, age=? where user_id=? 
Hibernate: select user0_.user_id as user_id, user0_.name as name, user0_.易做图 as 易做图, user0_.age as age from USER user0_ 
justin 
Age: 28 
Sex: M 
momor 
Age: 25 
Sex: F 
Bush 
Age: 25 
Sex: M 
Becky 
Age: 35 
Sex: F 

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