这里有一些关于JPA实体的讨论,以及应该为JPA实体类使用哪些hashCode()/equals()实现。它们中的大多数(如果不是全部)依赖于Hibernate,但是我想中立地讨论它们的jpa实现(顺便说一下,我使用的是EclipseLink)。

所有可能的实现都有其自身的优点和缺点:

hashCode()/equals()契约一致性(不可变性)用于列表/集操作 是否可以检测到相同的对象(例如来自不同会话的对象,来自惰性加载数据结构的动态代理) 实体在分离(或非持久化)状态下是否正确运行

在我看来,有三种选择:

Do not override them; rely on Object.equals() and Object.hashCode() hashCode()/equals() work cannot identify identical objects, problems with dynamic proxies no problems with detached entities Override them, based on the primary key hashCode()/equals() are broken correct identity (for all managed entities) problems with detached entities Override them, based on the Business-Id (non-primary key fields; what about foreign keys?) hashCode()/equals() are broken correct identity (for all managed entities) no problems with detached entities

我的问题是:

我是否错过了一个选择和/或赞成/反对的观点? 你选择了什么,为什么?

更新1:

通过“hashCode()/equals()是坏的”,我的意思是连续的hashCode()调用可能返回不同的值,这(当正确实现时)在对象API文档的意义上不是坏的,但是当试图从Map、Set或其他基于哈希的集合中检索更改的实体时,会导致问题。因此,JPA实现(至少是EclipseLink)在某些情况下不能正确工作。

更新2:

谢谢你的回答——大部分问题都很有质量。 不幸的是,我仍然不确定哪种方法最适合实际应用程序,或者如何确定最适合我的应用程序的方法。所以,我将保持这个问题的开放性,希望有更多的讨论和/或意见。


当前回答

我过去一直使用选项1,因为我知道这些讨论,并认为在我知道正确的事情之前最好什么都不做。这些系统仍在成功运行。

但是,下次我可能会尝试选项2 -使用数据库生成的Id。

如果未设置id, Hashcode和equals将抛出IllegalStateException。

这将防止涉及未保存实体的细微错误意外出现。

人们对这种方法有什么看法?

其他回答

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。

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.

我过去一直使用选项1,因为我知道这些讨论,并认为在我知道正确的事情之前最好什么都不做。这些系统仍在成功运行。

但是,下次我可能会尝试选项2 -使用数据库生成的Id。

如果未设置id, Hashcode和equals将抛出IllegalStateException。

这将防止涉及未保存实体的细微错误意外出现。

人们对这种方法有什么看法?