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。

其他回答

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?

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?

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)

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。

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();
    }

}