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[JSP] jsp Hibernate入门教程第1/3页

开发技术 开发技术 发布于:2021-10-24 15:23 | 阅读数:468 | 评论:0

例如:
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_.sex as sex, 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=?, sex=?, age=? where user_id=?
Hibernate: select user0_.user_id as user_id, user0_.name as name, user0_.sex as sex, 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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