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();
    }

}

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。

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。

虽然使用业务键(选项3)是最常推荐的方法(Hibernate社区wiki，“Java Persistence with Hibernate”第398页)，而且这是我们最常用的方法，但Hibernate有一个错误会破坏急于获取的集:HHH-3799。在这种情况下，Hibernate可以在字段初始化之前将一个实体添加到集合中。我不确定为什么这个错误没有得到更多的关注，因为它确实使推荐的业务键方法出现了问题。

我认为问题的核心是equals和hashCode应该基于不可变状态(参考Odersky等人)，而具有Hibernate管理的主键的Hibernate实体没有这样的不可变状态。当一个瞬态对象变成持久对象时，Hibernate会修改主键。当Hibernate在初始化过程中为对象补水时，业务键也会被Hibernate修改。

这就只剩下选项1了，基于对象身份继承java.lang.Object实现，或者使用James Brundege在“不要让Hibernate窃取你的身份”(Stijn Geukens的回答已经引用了)和Lance Arlaus在“对象生成:Hibernate集成的更好方法”中建议的应用程序管理的主键。

The biggest problem with option 1 is that detached instances can't be compared with persistent instances using .equals(). But that's OK; the contract of equals and hashCode leaves it up to the developer to decide what equality means for each class. So just let equals and hashCode inherit from Object. If you need to compare a detached instance to a persistent instance, you can create a new method explicitly for that purpose, perhaps boolean sameEntity or boolean dbEquivalent or boolean businessEquals.

实际上，似乎Option 2(主键)是最常用的。 自然的和不可变的业务密钥是很少的事情，创建和支持合成密钥对于解决情况来说太沉重了，这可能从来没有发生过。 看一下spring-data-jpa AbstractPersistable实现(唯一需要注意的是:对于Hibernate实现使用Hibernate. getclass)。

public boolean equals(Object obj) {
    if (null == obj) {
        return false;
    }
    if (this == obj) {
        return true;
    }
    if (!getClass().equals(ClassUtils.getUserClass(obj))) {
        return false;
    }
    AbstractPersistable<?> that = (AbstractPersistable<?>) obj;
    return null == this.getId() ? false : this.getId().equals(that.getId());
}

@Override
public int hashCode() {
    int hashCode = 17;
    hashCode += null == getId() ? 0 : getId().hashCode() * 31;
    return hashCode;
}

注意在HashSet/HashMap中操作新对象。 相反，选项1(保留对象实现)在合并后被破坏，这是非常常见的情况。

如果你没有业务键，并且需要在哈希结构中操作新实体，则将hashCode重写为常量，如下所示Vlad Mihalcea的建议。