MappedSuperclass - 在子类中更改SequenceGenerator [英] MappedSuperclass - Change SequenceGenerator in Subclass

问题描述

我正在使用JPA2和Hibernate，并尝试为我的实体引入一个公共基类。到目前为止它看起来像：

I'm using JPA2 with Hibernate and try to introduce a common base class for my entities. So far it looks like that:

@MappedSuperclass
public abstract class BaseEntity {

    @Id
    private Long id;

    @Override
    public int hashCode() {
        // ...
    }

    @Override
    public boolean equals(Object obj) {
        // ...
    }

    public Long getId() {
        return this.id;
    }

    public void setId(Long id) {
        this.id = id;
    }
}

但是，每个表都有一个序列 $ entityname_seq 我想用作序列生成器。如何从我的子类中设置它？我想我需要覆盖@GeneratedValue并使用@SequenceGenerator创建一个新的SequenceGenerator。

However, for every table theres a sequence $entityname_seq which I want to use as my sequence generator. How can I set that from my subclass? I think I need to override @GeneratedValue and create a new SequenceGenerator with @SequenceGenerator.

推荐答案

是的，这是可能的。您可以使用 @SequenceGenerator 注释覆盖默认生成器名称。

Yes, it is possible. You can override the default generator name with the @SequenceGenerator annotation.

• 基类

    @MappedSuperclass
    public abstract class PersistentEntity implements Serializable
    {
        private static final long serialVersionUID = 1L;

        @Id
        @GeneratedValue(strategy = GenerationType.SEQUENCE, generator = "default_gen")
        protected Long id = 0L;

        public Long getId()
        {
            return id;
        }

        public void setId(Long id)
        { 
            this.id = id;
        }
    }

• 序列（SQL）

  • Sequence (SQL)

    create sequence role_seq;

  
  

  • 派生类

  • Derived class

        @Entity
        @Table(name = "role")
        @SequenceGenerator(name = "default_gen", sequenceName = "role_seq", allocationSize = 1)
        public class Role extends PersistentEntity implements Serializable
        {
            private static final long serialVersionUID = 1L;
    
            @NotNull
            @Size(max = 32)
            private String name;
    
            public String getName()
            {
                 return name;
            }
    
            public void setName(String name)
            {
                 this.name = name;
            }   
        }

    
    
    

    
    
    • 这种方法在Hibernate 4.1中运行良好。 x，但它不在EclipseLink 2.x中。
    
    

    编辑

    
    
    • 根据评论，它似乎与EclipseLink 2.6.1-RC1一起使用。
    
    

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