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

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。

其他回答

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/

If you have a business key, then you should use that for equals and hashCode. If you don't have a business key, you should not leave it with the default Object equals and hashCode implementations because that does not work after you merge and entity. You can use the entity identifier in the equals method only if the hashCode implementation returns a constant value, like this: @Entity public class Book implements Identifiable<Long> { @Id @GeneratedValue private Long id; private String title; @Override public boolean equals(Object o) { if (this == o) return true; if (!(o instanceof Book)) return false; Book book = (Book) o; return getId() != null && Objects.equals(getId(), book.getId()); } @Override public int hashCode() { return getClass().hashCode(); } //Getters and setters omitted for brevity }

看看GitHub上的这个测试用例,它证明了这个解决方案很有魅力。

Jakarta Persistence 3.0,第4.12节写道:

相同抽象模式类型的两个实体当且仅当它们具有相同的主键值时相等。

我看不出为什么Java代码的行为应该有所不同。

If the entity class is in a so called "transient" state, i.e. it's not yet persisted and it has no identifier, then the hashCode/equals methods can not return a value, they ought to blow up, ideally implicitly with a NullPointerException when the method attempts to traverse the ID. Either way, this will effectively stop application code from putting a non-managed entity into a hash-based data structure. In fact, why not go one step further and blow up if the class and identifier are equal, but other important attributes such as the version are unequal (IllegalStateException)! Fail-fast in a deterministic way is always the preferred option.

警告:也要记录下爆发行为。文档本身很重要,但它也希望能够阻止初级开发人员在未来对您的代码做一些愚蠢的事情(他们倾向于压制发生NullPointerException的地方,他们最不关心的是副作用,lol)。

哦,总是使用getClass()而不是instanceof。equals方法要求对称性。如果b等于a,那么a必须等于b。对于子类,instanceof打破了这种关系(a不是b的实例)。

尽管我个人总是使用getClass(),即使在实现非实体类(类型是状态,所以子类添加状态,即使子类是空的或只包含行为),只有当类是final时,instanceof才会很好。但实体类必须不是最终的(§2.1),所以我们真的别无选择。

Some folks may not like getClass(), because of the persistence provider's proxy wrapping the object. This might have been a problem in the past, but it really shouldn't be. A provider not returning different proxy classes for different entities, well, I'd say that's not a very smart provider lol. Generally, we shouldn't solve a problem until there is a problem. And, it seems like Hibernate's own documentation doesn't even see it worthwhile mentioning. In fact, they elegantly use getClass() in their own examples (see this).

Lastly, if one has an entity subclass that is an entity, and the inheritance mapping strategy used is not the default ("single table"), but configured to be a "joined subtype", then the primary key in that subclass table will be the same as the superclass table. If the mapping strategy is "table per concrete class", then the primary key may be the same as in the superclass. An entity subclass is very likely to be adding state and therefore just as likely to be logically a different thing. But an equals implementation using instanceof can not necessarily and secondarily rely on the ID only, as we saw may be the same for different entities.

在我看来,instanceof在非final Java类中根本没有位置。对于持久实体来说尤其如此。

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()
}

我总是重写equals/hashcode,并基于业务id实现它。对我来说这是最合理的解决办法。请看下面的链接。

总而言之,这里列出了处理equals/hashCode的不同方法中哪些是有效的,哪些是无效的:

编辑:

为了解释为什么这对我有用:

I don't usually use hashed-based collection (HashMap/HashSet) in my JPA application. If I must, I prefer to create UniqueList solution. I think changing business id on runtime is not a best practice for any database application. On rare cases where there is no other solution, I'd do special treatment like remove the element and put it back to the hashed-based collection. For my model, I set the business id on constructor and doesn't provide setters for it. I let JPA implementation to change the field instead of the property. UUID solution seems to be overkill. Why UUID if you have natural business id? I would after all set the uniqueness of the business id in the database. Why having THREE indexes for each table in the database then?