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。

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; }

这是每个使用Java和JPA的IT系统中的常见问题。痛点不仅仅是实现equals()和hashCode()，它还影响组织引用实体的方式以及其客户机引用同一实体的方式。我已经看够了没有商业钥匙的痛苦，以至于我写了自己的博客来表达我的观点。

简而言之:使用一个简短的、人类可读的、带有有意义前缀的顺序ID作为业务键，该ID生成时不依赖于RAM以外的任何存储。Twitter的雪花就是一个很好的例子。

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()
}

我们通常在实体中有两个id:

仅用于持久化层(以便持久化提供程序和数据库能够找出对象之间的关系)。 是为了我们的应用程序需要(特别是equals()和hashCode())

来看看:

@Entity
public class User {

    @Id
    private int id;  // Persistence ID
    private UUID uuid; // Business ID

    // assuming all fields are subject to change
    // If we forbid users change their email or screenName we can use these
    // fields for business ID instead, but generally that's not the case
    private String screenName;
    private String email;

    // I don't put UUID generation in constructor for performance reasons. 
    // I call setUuid() when I create a new entity
    public User() {
    }

    // This method is only called when a brand new entity is added to 
    // persistence context - I add it as a safety net only but it might work 
    // for you. In some cases (say, when I add this entity to some set before 
    // calling em.persist()) setting a UUID might be too late. If I get a log 
    // output it means that I forgot to call setUuid() somewhere.
    @PrePersist
    public void ensureUuid() {
        if (getUuid() == null) {
            log.warn(format("User's UUID wasn't set on time. " 
                + "uuid: %s, name: %s, email: %s",
                getUuid(), getScreenName(), getEmail()));
            setUuid(UUID.randomUUID());
        }
    }

    // equals() and hashCode() rely on non-changing data only. Thus we 
    // guarantee that no matter how field values are changed we won't 
    // lose our entity in hash-based Sets.
    @Override
    public int hashCode() {
        return getUuid().hashCode();
    }

    // Note that I don't use direct field access inside my entity classes and
    // call getters instead. That's because Persistence provider (PP) might
    // want to load entity data lazily. And I don't use 
    //    this.getClass() == other.getClass() 
    // for the same reason. In order to support laziness PP might need to wrap
    // my entity object in some kind of proxy, i.e. subclassing it.
    @Override
    public boolean equals(final Object obj) {
        if (this == obj)
            return true;
        if (!(obj instanceof User))
            return false;
        return getUuid().equals(((User) obj).getUuid());
    }

    // Getters and setters follow
}

编辑:澄清我关于调用setUuid()方法的观点。下面是一个典型的场景:

User user = new User();
// user.setUuid(UUID.randomUUID()); // I should have called it here
user.setName("Master Yoda");
user.setEmail("yoda@jedicouncil.org");

jediSet.add(user); // here's bug - we forgot to set UUID and 
                   //we won't find Yoda in Jedi set

em.persist(user); // ensureUuid() was called and printed the log for me.

jediCouncilSet.add(user); // Ok, we got a UUID now

当我运行测试并看到日志输出时，我解决了这个问题:

User user = new User();
user.setUuid(UUID.randomUUID());

或者，也可以提供一个单独的构造函数:

@Entity
public class User {

    @Id
    private int id;  // Persistence ID
    private UUID uuid; // Business ID

    ... // fields

    // Constructor for Persistence provider to use
    public User() {
    }

    // Constructor I use when creating new entities
    public User(UUID uuid) {
        setUuid(uuid);
    }

    ... // rest of the entity.
}

我的例子是这样的:

User user = new User(UUID.randomUUID());
...
jediSet.add(user); // no bug this time

em.persist(user); // and no log output

我使用默认构造函数和setter，但您可能会发现双构造函数方法更适合您。