jsp Hibernate入门教程第1/3页

Java  /  管理员 发布于 7年前   153

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_.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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