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。

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?

我个人已经在不同的项目中使用了这三种策略。我必须说，选项1在我看来是现实应用中最可行的。以我的经验来看，打破hashCode()/equals()一致性会导致许多疯狂的错误，因为你每次都会遇到这样的情况:在一个实体被添加到一个集合后，相等的结果发生了变化。

但也有更多的选择(也有它们的优点和缺点):

a) hashCode/equals基于一组不可变的、非空的、构造函数赋值的字段

(+)三个标准都有保证

(-)字段值必须可用以创建新实例

(-)如果你必须改变其中一个，处理起来会很复杂

b) hashCode/equals基于应用程序(在构造函数中)分配的主键，而不是JPA

(+)三个标准都有保证

(-)您不能利用简单可靠的ID生成策略，如DB序列

(-)如果在分布式环境(客户端/服务器)或应用服务器集群中创建新实体会很复杂

c) hashCode/equals基于实体的构造函数分配的UUID

(+)三个标准都有保证

(-)生成UUID的开销

(-)可能会有使用两次相同UUID的风险，这取决于所使用的算法(可能由DB上的唯一索引检测到)

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。

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/