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。

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; }

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)

如果UUID是许多人的答案，为什么我们不使用业务层的工厂方法来创建实体并在创建时分配主键呢?

例如:

@ManagedBean
public class MyCarFacade {
  public Car createCar(){
    Car car = new Car();
    em.persist(car);
    return car;
  }
}

通过这种方式，我们可以从持久化提供程序获得实体的默认主键，并且我们的hashCode()和equals()函数可以依赖于它。

我们还可以声明Car的构造函数受保护，然后在业务方法中使用反射来访问它们。这样，开发人员就不会打算用new实例化Car，而是通过factory方法。

来说,如何?

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。