如何在JPA中生成自定义ID [英] How to generate Custom Id in JPA
本文介绍了如何在JPA中生成自定义ID的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我想在JPA中生成自定义ID,它必须是表的主键. 有很多使用休眠模式创建自定义ID的示例,例如
i want generate Custom Id in JPA it must be primary key of table.
there are many examples to create Custom Id using hibernate like this
i want same implementation but in JPA.The id must be alphanumeric like STAND0001
谢谢.
推荐答案
您可以使用GenericGenerator
来做到这一点:
You can do it using GenericGenerator
like this :
@Entity
public class Client {
@Id
@GenericGenerator(name = "client_id", strategy = "com.eframe.model.generator.ClientIdGenerator")
@GeneratedValue(generator = "client_id")
@Column(name="client_id")
private String clientId;
}
和自定义生成器类(将在ID前面添加前缀,您可以根据自己的喜好进行设置):
and the custom generator class (will add prefix to the ID, you can make it do what you like):
public class ClientIdGenerator implements IdentifierGenerator {
@Override
public Serializable generate(SessionImplementor session, Object object)
throws HibernateException {
String prefix = "cli";
Connection connection = session.connection();
try {
Statement statement=connection.createStatement();
ResultSet rs=statement.executeQuery("select count(client_id) as Id from Client");
if(rs.next())
{
int id=rs.getInt(1)+101;
String generatedId = prefix + new Integer(id).toString();
return generatedId;
}
} catch (SQLException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
return null;
}
}
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