更新、刪除數據

如果您是在同一個Session中取出數據並想要馬上進行更新，則只要先查詢並取出對象，通過setXXX()方法設定好新的值，然後調用session.flush()即可在同一個Session中更新指定的數據，例如：
代碼:

import com.xtedu.teach.hibernate.mappings.*;
import org.hibernate.*;
import org.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());
}
}
}

這個程序會顯示數據表中的所有數據，並將數據表中的第一筆數據更新，一個執行的結果如下：
代碼:

log4j:WARN No appenders could be found for logger (org.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

 

如果您打開了一個Session，從數據表中獲取了數據，之後關閉Session，當使用者修改完畢並進行存儲時，您要重新打開一個Session，使用update()方法將對象中的數據更新至對應的數據表中，一個例子如下：
代碼:

import com.xtedu.teach.hibernate.mappings.*;
import org.hibernate.*;
import org.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");
// 關閉這個Session
session.close();
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());
}
// 使用者作一些操作，之後存儲
updated.setName("caterpillar");
// 開啓一個新的Session
session = sessionFactory.openSession();
// 更新數據
session.update(updated);
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() +
"
'5cn\tSex: " + user.getSex());
}
}
}

這個程序執行的結果範例如下，您可以看看實際上執行了哪些SQL：
代碼:

log4j:WARN No appenders could be found for logger (org.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_
justin
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_
caterpillar
Age: 28
Sex: M
momor
Age: 25
Sex: F
Bush
Age: 25
Sex: M
Becky
Age: 35
Sex: F

Hibernate提供了一個saveOrUpdate()方法，爲數據的存儲或更新提供了一個統一的操作接口，藉由定義映射文件時，設定<id>標籤的unsaved-value來決定什麼是新的值必需插入，什麼是已有的值必須更新：
代碼:

<id name="id" type="string" unsaved-value="null">
<column name="user_id" sql-type="char(32)" />
<generator class="uuid.hex"/>
</id>

unsaved-value可以設定的值包括：
*any - 總是存儲
*none - 總是更新
*null - id爲null時存儲（預設）
*valid - id爲null或是指定值時存儲
這樣設定之後，您可以使用session.saveOrUpdate(updated);來取代上一個程序的session.update(updated);方法。
如果要刪除數據，只要使用delete()方法即可，直接看個例子。
代碼:

import com.xtedu.teach.hibernate.mappings.*;
import org.hibernate.*;
import org.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());
}
session.delete(updated);
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());
}
}
}

一個執行的結果範例如下：
代碼:

log4j:WARN No appenders could be found for logger (org.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_
justin
Age: 28
Sex: M
momor
Age: 25
Sex: F
Bush
Age: 25
Sex: M
Becky
Age: 35
Sex: F
Hibernate: delete from USER 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_
momor
Age: 25
Sex: F
Bush
Age: 25
Sex: M
Becky
Age: 35
Sex: F

Hibernate對於數據的更新、刪除等動作，是依懶id值來判定，如果您已知id值，則可以使用load()方法來載入數據，例如：
代碼:

User user = (User) session.load(User.class, id);