java - 如何使用 JPA 持久化两个实体

Question

我在我的 web 应用程序中使用 JPA，但我不知道如何保留两个相互关联的新实体。这里有一个例子:

这是两个实体

+-----------------+   +--------------------+|     Consumer    |   |   ProfilePicture   |+-----------------+   +--------------------+| id    (PK)      |---| consumerId (PPK+FK)|| userName        |   | url                |+-----------------+   +--------------------+

The Consumer has an id and some other values. The ProfilePicture uses the Consumer's id as it's own primary key and as foreign key. (Since a ProfilePicture will not exist without a Consumer and not every Consumer has a ProfilePicture)

I used NetBeans to generate the entity classes and the session beans (facades).

This is how they look like in short

Consumer.java

@Entity
@Table(name = "Consumer")
@NamedQueries({...})
public class Consumer implements Serializable {

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    @Basic(optional = false)
    @Column(name = "id")
    private Integer id;

    @Basic(optional = false)
    @NotNull
    @Size(min = 1, max = 50)
    @Column(name = "userName")
    private String userName;     

    @OneToOne(cascade = CascadeType.ALL, mappedBy = "consumer")
    private ProfilePicture profilePicture;

    /* and all the basic getters and setters */
    (...)
}

ProfilePicture.java

@Entity
@Table(name = "ProfilePicture")
@XmlRootElement
@NamedQueries({...})
public class ProfilePicture implements Serializable {

    @Id
    @Basic(optional = false)
    @NotNull
    @Column(name = "consumerId")
    private Integer consumerId;

    @Basic(optional = false)
    @NotNull
    @Size(min = 1, max = 255)
    @Column(name = "url")
    private String url;

    @JoinColumn(name = "consumerId", referencedColumnName = "id", insertable = false, updatable = false)
    @OneToOne(optional = false)
    private Consumer consumer;

    /* and all the basic getters and setters */
    (...)
}

所以当我想用他的 ProfilePicture 创建一个 Consumer 时，我想我会这样做:

   ProfilePicture profilePicture = new ProfilePicture("http://www.url.to/picture.jpg");  // create the picture object
   Consumer consumer = new Consumer("John Doe"); // create the consumer object

   profilePicture.setConsumer(consumer);        // set the consumer in the picture (so JPA can take care about the relation      
   consumerFacade.create(consumer);             // the facade classes to persist the consumer
   profilePictureFacade.create(profilePicture);  // and when the consumer is persisted (and has an id) persist the picture

我的问题

我几乎尝试了每种组合中的所有方法，但 JPA 似乎无法单独链接这两个实体。大多数时候我会收到这样的错误:

 EJB5184:A system exception occurred during an invocation on EJB ConsumerFacade, method: public void com.me.db.resources.bean.ConsumerFacade.create(com.mintano.backendclientserver.db.resources.entity.Consumer)
 (...) 
Bean Validation constraint(s) violated while executing Automatic Bean Validation on callback event:'prePersist'. Please refer to embedded ConstraintViolations for details.

据我所知，这是因为 ProfilePicture 不知道 Consumer 的 ID，因此实体无法持久化。

唯一可行的方法是，首先保留 Consumer，将其 id 设置为 ProfilePicture，然后保留图片:

   ProfilePicture profilePicture = new ProfilePicture("http://www.url.to/picture.jpg");  // create the picture object
   Consumer consumer = new Consumer("John Doe"); // create the consumer object

   consumerFacade.create(consumer);             // the facade classes to persist the consumer
   profilePicture.setConsumerId(consumer.getId()); // set the consumer's new id in the picture     

   profilePictureFacade.create(profilePicture);  // and when the consumer is persisted (and has an id) persist the picture

然而，这两个表只是一个示例，自然数据库要复杂得多，像这样手动设置 id 似乎非常不灵活，我担心事情会过于复杂。特别是因为我不能在一个事务中保留所有实体(这看起来效率很低)。

我做得对吗？或者还有其他更标准的方法吗？

编辑:我的解决方案

作为FTR建议，一个问题是 ProfilePicture 表缺少 id(我使用 Consumer.id 作为外部和主要)。

表格现在看起来像这样:

+-----------------+   +--------------------+|     Consumer    |   |   ProfilePicture   |+-----------------+   +--------------------+| id    (PK)      |_  | id (PK)            || userName        | \_| consumerId (FK)    |+-----------------+   | url                |                      +--------------------+

Then Alan Hay told me to Always encapsulate add/remove to relationships and then you can ensure correctness, which I did:

Consumer.java

public void addProfilePicture(ProfilePicture profilePicture) {
    profilePicture.setConsumerId(this);  
    if (profilePictureCollection == null) {
        this.profilePictureCollection = new ArrayList<>();
    }
    this.profilePictureCollection.add(profilePicture);
}

由于 ProfilePicture 现在有自己的 id，它变成了 OneToMany 关系，所以每个 Consumer 现在可以有很多个人资料图片。这不是我最初的意图，但我可以忍受它:) 因此我不能只为消费者设置一个 ProfilePicture，而是必须将它添加到图片集合中(如上所述)。

这是我实现的唯一附加方法，现在可以使用了。再次感谢您的帮助!

Answer 1

当持久化关系的非拥有方的实例时(包含“mappedBy”，在您的情况下是消费者)，您必须始终确保关系的双方都设置为具有预期的级联工作。

当然，您应该始终这样做，以确保您的领域模型是正确的。

Consumer c = new Consumer();
ProfilePicure p = new ProfilePicture();
c.setProfilePicture(p);//see implementation
//persist c

消费者.java

    @Entity
    @Table(name = "Consumer")
    @NamedQueries({...})
    public class Consumer implements Serializable {

        @Id
        @GeneratedValue(strategy = GenerationType.IDENTITY)
        @Basic(optional = false)
        @Column(name = "id")
        private Integer id;

        @Basic(optional = false)
        @NotNull
        @Size(min = 1, max = 50)
        @Column(name = "userName")
        private String userName;     

        @OneToOne(cascade = CascadeType.ALL, mappedBy = "consumer")
        private ProfilePicture profilePicture;

        public void setProfilePicture(ProfilePicture profilePicture){
            //SET BOTH SIDES OF THE RELATIONSHIP
            this.profilePicture = profilePicture;
            profilePicture.setConsumer(this);
        }
}

始终封装添加/删除关系，然后您可以确保正确性:

public class Parent{
private Set<Child> children;

public Set<Child> getChildren(){
    return Collections.unmodifiableSet(children); //no direct access:force clients to use add/remove methods
}

public void addChild(Child child){
    child.setParent(this); 
    children.add(child);
}

public class Child(){
    private Parent parent;
}