这里有一些关于JPA实体的讨论,以及应该为JPA实体类使用哪些hashCode()/equals()实现。它们中的大多数(如果不是全部)依赖于Hibernate,但是我想中立地讨论它们的jpa实现(顺便说一下,我使用的是EclipseLink)。

所有可能的实现都有其自身的优点和缺点:

hashCode()/equals()契约一致性(不可变性)用于列表/集操作 是否可以检测到相同的对象(例如来自不同会话的对象,来自惰性加载数据结构的动态代理) 实体在分离(或非持久化)状态下是否正确运行

在我看来,有三种选择:

Do not override them; rely on Object.equals() and Object.hashCode() hashCode()/equals() work cannot identify identical objects, problems with dynamic proxies no problems with detached entities Override them, based on the primary key hashCode()/equals() are broken correct identity (for all managed entities) problems with detached entities Override them, based on the Business-Id (non-primary key fields; what about foreign keys?) hashCode()/equals() are broken correct identity (for all managed entities) no problems with detached entities

我的问题是:

我是否错过了一个选择和/或赞成/反对的观点? 你选择了什么,为什么?

更新1:

通过“hashCode()/equals()是坏的”,我的意思是连续的hashCode()调用可能返回不同的值,这(当正确实现时)在对象API文档的意义上不是坏的,但是当试图从Map、Set或其他基于哈希的集合中检索更改的实体时,会导致问题。因此,JPA实现(至少是EclipseLink)在某些情况下不能正确工作。

更新2:

谢谢你的回答——大部分问题都很有质量。 不幸的是,我仍然不确定哪种方法最适合实际应用程序,或者如何确定最适合我的应用程序的方法。所以,我将保持这个问题的开放性,希望有更多的讨论和/或意见。


当前回答

如果你想对你的set使用equals()/hashCode(),也就是说同一个实体只能出现一次,那么只有一个选项:选项2。这是因为根据定义,实体的主键永远不会改变(如果有人确实更新了它,它就不再是同一个实体了)

您应该从字面上理解:因为equals()/hashCode()是基于主键的,所以在设置主键之前,您不能使用这些方法。所以你不应该把实体放到集合里,直到它们被赋主键。(是的,uuid和类似的概念可能有助于早期分配主键。)

Now, it's theoretically also possible to achieve that with Option 3, even though so-called "business-keys" have the nasty drawback that they can change: "All you'll have to do is delete the already inserted entities from the set(s), and re-insert them." That is true - but it also means, that in a distributed system, you'll have to make sure, that this is done absolutely everywhere the data has been inserted to (and you'll have to make sure, that the update is performed, before other things occur). You'll need a sophisticated update mechanism, especially if some remote systems aren't currently reachable...

只有当集合中的所有对象都来自同一个Hibernate会话时,才可以使用选项1。Hibernate文档在13.1.3章中非常清楚地说明了这一点。考虑对象同一性:

Within a Session the application can safely use == to compare objects. However, an application that uses == outside of a Session might produce unexpected results. This might occur even in some unexpected places. For example, if you put two detached instances into the same Set, both might have the same database identity (i.e., they represent the same row). JVM identity, however, is by definition not guaranteed for instances in a detached state. The developer has to override the equals() and hashCode() methods in persistent classes and implement their own notion of object equality.

它继续主张选择3:

这里有一个警告:永远不要使用数据库标识符来实现相等。使用由唯一的、通常是不可变的属性组合而成的业务键。如果将瞬态对象持久化,则数据库标识符将更改。如果瞬态实例(通常与分离实例一起)保存在Set中,更改hashcode将破坏Set的契约。

这是真的,如果你

不能提前分配id(例如使用uuid) 当对象处于瞬态时,你肯定想把它们放到集合中。

否则,您可以自由选择选项2。

然后它提到了相对稳定性的需求:

业务键的属性不必像数据库主键那样稳定;只要对象在同一集合中,你就必须保证稳定性。

这是正确的。我所看到的实际问题是:如果你不能保证绝对的稳定性,你如何能够保证“只要对象在同一个集合中”的稳定性。我可以想象一些特殊的情况(比如只在对话中使用集合,然后将其丢弃),但我会质疑这种方法的一般实用性。


短版:

选项1只能用于单个会话中的对象。 如果可以,使用选项2。(尽早分配PK,因为在分配PK之前你不能在集合中使用对象。) 如果你能保证相对的稳定性,你可以使用选项3。但是要小心。

其他回答

我试着自己回答这个问题,直到我读了这篇文章,尤其是画了一个,我才完全满意找到的解决方案。我喜欢他懒创建UUID和最佳存储它的方式。

但我想增加更多的灵活性,即惰性创建UUID仅当hashCode()/equals()被访问时,第一次持久化实体与每个解决方案的优点:

Equals()表示“对象指向相同的逻辑实体” 尽可能使用数据库ID,因为为什么我要做两次工作(性能问题) 防止在尚未持久的实体上访问hashCode()/equals()时出现问题,并在它确实被持久后保持相同的行为

我真的很感激对我的混合解决方案的反馈如下

public class MyEntity { @Id() @Column(name = "ID", length = 20, nullable = false, unique = true) @GeneratedValue(strategy = GenerationType.IDENTITY) private Long id = null; @Transient private UUID uuid = null; @Column(name = "UUID_MOST", nullable = true, unique = false, updatable = false) private Long uuidMostSignificantBits = null; @Column(name = "UUID_LEAST", nullable = true, unique = false, updatable = false) private Long uuidLeastSignificantBits = null; @Override public final int hashCode() { return this.getUuid().hashCode(); } @Override public final boolean equals(Object toBeCompared) { if(this == toBeCompared) { return true; } if(toBeCompared == null) { return false; } if(!this.getClass().isInstance(toBeCompared)) { return false; } return this.getUuid().equals(((MyEntity)toBeCompared).getUuid()); } public final UUID getUuid() { // UUID already accessed on this physical object if(this.uuid != null) { return this.uuid; } // UUID one day generated on this entity before it was persisted if(this.uuidMostSignificantBits != null) { this.uuid = new UUID(this.uuidMostSignificantBits, this.uuidLeastSignificantBits); // UUID never generated on this entity before it was persisted } else if(this.getId() != null) { this.uuid = new UUID(this.getId(), this.getId()); // UUID never accessed on this not yet persisted entity } else { this.setUuid(UUID.randomUUID()); } return this.uuid; } private void setUuid(UUID uuid) { if(uuid == null) { return; } // For the one hypothetical case where generated UUID could colude with UUID build from IDs if(uuid.getMostSignificantBits() == uuid.getLeastSignificantBits()) { throw new Exception("UUID: " + this.getUuid() + " format is only for internal use"); } this.uuidMostSignificantBits = uuid.getMostSignificantBits(); this.uuidLeastSignificantBits = uuid.getLeastSignificantBits(); this.uuid = uuid; }

显然,这里已经有了非常有用的答案,但我将告诉你我们是怎么做的。

我们什么也不做。

如果我们确实需要= /hashcode来处理集合,则使用uuid。 您只需在构造函数中创建UUID。我们使用http://wiki.fasterxml.com/JugHome作为UUID。UUID的CPU开销稍高,但与序列化和db访问相比便宜。

如果你想对你的set使用equals()/hashCode(),也就是说同一个实体只能出现一次,那么只有一个选项:选项2。这是因为根据定义,实体的主键永远不会改变(如果有人确实更新了它,它就不再是同一个实体了)

您应该从字面上理解:因为equals()/hashCode()是基于主键的,所以在设置主键之前,您不能使用这些方法。所以你不应该把实体放到集合里,直到它们被赋主键。(是的,uuid和类似的概念可能有助于早期分配主键。)

Now, it's theoretically also possible to achieve that with Option 3, even though so-called "business-keys" have the nasty drawback that they can change: "All you'll have to do is delete the already inserted entities from the set(s), and re-insert them." That is true - but it also means, that in a distributed system, you'll have to make sure, that this is done absolutely everywhere the data has been inserted to (and you'll have to make sure, that the update is performed, before other things occur). You'll need a sophisticated update mechanism, especially if some remote systems aren't currently reachable...

只有当集合中的所有对象都来自同一个Hibernate会话时,才可以使用选项1。Hibernate文档在13.1.3章中非常清楚地说明了这一点。考虑对象同一性:

Within a Session the application can safely use == to compare objects. However, an application that uses == outside of a Session might produce unexpected results. This might occur even in some unexpected places. For example, if you put two detached instances into the same Set, both might have the same database identity (i.e., they represent the same row). JVM identity, however, is by definition not guaranteed for instances in a detached state. The developer has to override the equals() and hashCode() methods in persistent classes and implement their own notion of object equality.

它继续主张选择3:

这里有一个警告:永远不要使用数据库标识符来实现相等。使用由唯一的、通常是不可变的属性组合而成的业务键。如果将瞬态对象持久化,则数据库标识符将更改。如果瞬态实例(通常与分离实例一起)保存在Set中,更改hashcode将破坏Set的契约。

这是真的,如果你

不能提前分配id(例如使用uuid) 当对象处于瞬态时,你肯定想把它们放到集合中。

否则,您可以自由选择选项2。

然后它提到了相对稳定性的需求:

业务键的属性不必像数据库主键那样稳定;只要对象在同一集合中,你就必须保证稳定性。

这是正确的。我所看到的实际问题是:如果你不能保证绝对的稳定性,你如何能够保证“只要对象在同一个集合中”的稳定性。我可以想象一些特殊的情况(比如只在对话中使用集合,然后将其丢弃),但我会质疑这种方法的一般实用性。


短版:

选项1只能用于单个会话中的对象。 如果可以,使用选项2。(尽早分配PK,因为在分配PK之前你不能在集合中使用对象。) 如果你能保证相对的稳定性,你可以使用选项3。但是要小心。

我使用类EntityBase和继承到我所有的JPA实体,这对我来说非常好。

/**
 * @author marcos.oliveira
 */
@MappedSuperclass
public abstract class EntityBase<TId extends Serializable> implements Serializable{
    /**
     *
     */
    private static final long serialVersionUID = 1L;

    @Id
    @Column(name = "id", unique = true, nullable = false)
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    protected TId id;



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

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

    @Override
    public int hashCode() {
        return (super.hashCode() * 907) + Objects.hashCode(getId());//this.getId().hashCode();
    }

    @Override
    public String toString() {
        return super.toString() + " [Id=" + id + "]";
    }

    @Override
    public boolean equals(Object obj) {
        if (this == obj) {
            return true;
        }
        if (obj == null || getClass() != obj.getClass()) {
            return false;
        }
        EntityBase entity = (EntityBase) obj;
        if (entity.id == null || id == null) {
            return false;
        }
        return Objects.equals(id, entity.id);
    }
}

参考:https://thorben-janssen.com/ultimate-guide-to-implementing-equals-and-hashcode-with-hibernate/

下面是一个简单的(经过测试的)Scala解决方案。

请注意,此解决方案不属于这3类中的任何一类 在问题中给出。 我所有的实体都是UUIDEntity的子类,所以我遵循 不要重复自己(DRY)原则。 如果需要,可以使UUID生成更精确(通过使用更多 伪随机数)。

Scala代码:

import javax.persistence._
import scala.util.Random

@Entity
@Inheritance(strategy = InheritanceType.TABLE_PER_CLASS)
abstract class UUIDEntity {
  @Id  @GeneratedValue(strategy = GenerationType.TABLE)
  var id:java.lang.Long=null
  var uuid:java.lang.Long=Random.nextLong()
  override def equals(o:Any):Boolean= 
    o match{
      case o : UUIDEntity => o.uuid==uuid
      case _ => false
    }
  override def hashCode() = uuid.hashCode()
}