阅读这篇关于主题的非常好的文章:不要让Hibernate窃取您的身份。

文章的结论是这样的:

Object identity is deceptively hard to implement correctly when objects are persisted to a database. However, the problems stem entirely from allowing objects to exist without an id before they are saved. We can solve these problems by taking the responsibility of assigning object IDs away from object-relational mapping frameworks such as Hibernate. Instead, object IDs can be assigned as soon as the object is instantiated. This makes object identity simple and error-free, and reduces the amount of code needed in the domain model.

我同意Andrew的回答。我们在应用程序中做同样的事情，但不是将uuid存储为VARCHAR/CHAR，而是将其分割为两个长值。请参阅UUID.getLeastSignificantBits()和UUID.getMostSignificantBits()。

还有一件事需要考虑，对UUID. randomuuid()的调用非常慢，因此您可能希望只在需要时才惰性地生成UUID，例如在持久化期间或调用equals()/hashCode()期间

@MappedSuperclass
public abstract class AbstractJpaEntity extends AbstractMutable implements Identifiable, Modifiable {

    private static final long   serialVersionUID    = 1L;

    @Version
    @Column(name = "version", nullable = false)
    private int                 version             = 0;

    @Column(name = "uuid_least_sig_bits")
    private long                uuidLeastSigBits    = 0;

    @Column(name = "uuid_most_sig_bits")
    private long                uuidMostSigBits     = 0;

    private transient int       hashCode            = 0;

    public AbstractJpaEntity() {
        //
    }

    public abstract Integer getId();

    public abstract void setId(final Integer id);

    public boolean isPersisted() {
        return getId() != null;
    }

    public int getVersion() {
        return version;
    }

    //calling UUID.randomUUID() is pretty expensive, 
    //so this is to lazily initialize uuid bits.
    private void initUUID() {
        final UUID uuid = UUID.randomUUID();
        uuidLeastSigBits = uuid.getLeastSignificantBits();
        uuidMostSigBits = uuid.getMostSignificantBits();
    }

    public long getUuidLeastSigBits() {
        //its safe to assume uuidMostSigBits of a valid UUID is never zero
        if (uuidMostSigBits == 0) {
            initUUID();
        }
        return uuidLeastSigBits;
    }

    public long getUuidMostSigBits() {
        //its safe to assume uuidMostSigBits of a valid UUID is never zero
        if (uuidMostSigBits == 0) {
            initUUID();
        }
        return uuidMostSigBits;
    }

    public UUID getUuid() {
        return new UUID(getUuidMostSigBits(), getUuidLeastSigBits());
    }

    @Override
    public int hashCode() {
        if (hashCode == 0) {
            hashCode = (int) (getUuidMostSigBits() >> 32 ^ getUuidMostSigBits() ^ getUuidLeastSigBits() >> 32 ^ getUuidLeastSigBits());
        }
        return hashCode;
    }

    @Override
    public boolean equals(final Object obj) {
        if (obj == null) {
            return false;
        }
        if (!(obj instanceof AbstractJpaEntity)) {
            return false;
        }
        //UUID guarantees a pretty good uniqueness factor across distributed systems, so we can safely
        //dismiss getClass().equals(obj.getClass()) here since the chance of two different objects (even 
        //if they have different types) having the same UUID is astronomical
        final AbstractJpaEntity entity = (AbstractJpaEntity) obj;
        return getUuidMostSigBits() == entity.getUuidMostSigBits() && getUuidLeastSigBits() == entity.getUuidLeastSigBits();
    }

    @PrePersist
    public void prePersist() {
        // make sure the uuid is set before persisting
        getUuidLeastSigBits();
    }

}

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。

业务密钥方法不适合我们。我们使用DB生成的ID、临时临时tempId和重写equal()/hashcode()来解决这个困境。所有实体都是Entity的后代。优点:

DB中没有额外字段 在后代实体中没有额外的编码，一种方法适用于所有的实体 没有性能问题(如UUID)， DB Id生成 使用hashmap没有问题(不需要记住equal & etc的使用)。 新实体的Hashcode即使在持久化后也不会及时更改

缺点:

序列化和反序列化非持久化实体可能会出现问题 从DB重新加载后，保存的实体的Hashcode可能会改变 非持久化对象被认为总是不同的(也许这是对的?) 还有什么?

看看我们的代码:

@MappedSuperclass
abstract public class Entity implements Serializable {

    @Id
    @GeneratedValue
    @Column(nullable = false, updatable = false)
    protected Long id;

    @Transient
    private Long tempId;

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

    public Long getId() {
        return id;
    }

    private void setTempId(Long tempId) {
        this.tempId = tempId;
    }

    // Fix Id on first call from equal() or hashCode()
    private Long getTempId() {
        if (tempId == null)
            // if we have id already, use it, else use 0
            setTempId(getId() == null ? 0 : getId());
        return tempId;
    }

    @Override
    public boolean equals(Object obj) {
        if (super.equals(obj))
            return true;
        // take proxied object into account
        if (obj == null || !Hibernate.getClass(obj).equals(this.getClass()))
            return false;
        Entity o = (Entity) obj;
        return getTempId() != 0 && o.getTempId() != 0 && getTempId().equals(o.getTempId());
    }

    // hash doesn't change in time
    @Override
    public int hashCode() {
        return getTempId() == 0 ? super.hashCode() : getTempId().hashCode();
    }
}

阅读这篇关于主题的非常好的文章:不要让Hibernate窃取您的身份。

文章的结论是这样的:

Object identity is deceptively hard to implement correctly when objects are persisted to a database. However, the problems stem entirely from allowing objects to exist without an id before they are saved. We can solve these problems by taking the responsibility of assigning object IDs away from object-relational mapping frameworks such as Hibernate. Instead, object IDs can be assigned as soon as the object is instantiated. This makes object identity simple and error-free, and reduces the amount of code needed in the domain model.

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/