java - 在 Spring JPA 中是否有更简洁的方式构建 MappedSuperclass Tree 类？

Question

我目前有几个充当树的实体，需要将它们保存到数据库中。

所以为了不重复代码，我构建了这个类:

@MappedSuperclass
public abstract class TreeStructure<T extends TreeStructure>
{
    @ManyToOne(cascade = CascadeType.PERSIST)
    private T  parent;

    @OneToMany(mappedBy = "parent", fetch = FetchType.LAZY, cascade = CascadeType.PERSIST)
    protected Set<T> children = new HashSet<>();

    /**
     * Function that is used before deleting this entity. It joins this.children to this.parent and viceversa.
     */
    @Transactional
    @PreRemove
    public void preDelete()
    {
        unregisterInParentsChildren();

        while (!children.isEmpty())
        {
            children.iterator().next().setParent(parent);
        }

    }

    public abstract long getId();

    protected void setParent(T pParent)
    {
        unregisterInParentsChildren();
        parent = pParent;
        registerInParentsChildren();
    }

    /**
     * Register this TreeStructure in the child list of its parent if it's not null.
     */
    private void registerInParentsChildren()
    {
        getParent().ifPresent((pParent) -> pParent.children.add(this));
    }

    /**
     * Unregister this TreeStructure in the child list of its parent if it's not null.
     */
    private void unregisterInParentsChildren()
    {
        getParent().ifPresent((pParent) -> pParent.children.remove(this));
    }

    /**
     * Move this TreeStructure to an new parent TreeStructure.
     *
     * @param pNewParent the new parent
     */
    public void move(final T pNewParent)
    {
        if (pNewParent == null)
        {
            throw new IllegalArgumentException("New Parent required");
        }

        if (!isProperMoveTarget(pNewParent) /* detect circles... */)
        {
            throw new IllegalArgumentException(String.format("Unable to move Object %1$s to new Object Parent %2$s", getId(), pNewParent.getId()));
        }

        setParent(pNewParent);
    }

    private boolean isProperMoveTarget(TreeStructure pParent)
    {
        if (pParent == null)
        {
            return true;
        }
        if (pParent == this)
        {
            return false;
        }

        return isProperMoveTarget(pParent.parent);
    }

    public int getLevel()
    {
        return getParent().map(pParent -> pParent.getLevel() + 1).orElse(1);
    }

    /**
     * Return the <strong>unmodifiable</strong> children of this TreeStructure.
     *
     * @return the child nodes.
     */
    public Set<T> getChildren()
    {
        return Collections.unmodifiableSet(this.children);
    }

    public Optional<T> getParent()
    {
        return Optional.ofNullable(parent);
    }

    public Optional<Long> getParentCategoryId()
    {
        return parent == null ? Optional.empty() : Optional.of(parent.getId());
    }
}

然后要实际实现它，我只需执行以下操作:

@Entity(name = "CATEGORY")
public class Category extends TreeStructure<Category>
{
    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    @JsonProperty("category_id")
    private long id;

// etc...

据我所知，一切都很顺利，但每次我进入 TreeStructure 类 Intellij 都会突出显示一些错误:

mappedBy = "parent" -> Cannot resolve attribute parent.

children.iterator().next().setParent(parent) -> Unchecked call to setParent(T) as a member of raw type TreeStructure

pParent.children.add(this) -> Unchecked call to add(E) as a member of raw type java.util.Set

我也尝试不使用泛型，所以我可以只使用抽象 TreeStructure，然后从其他类扩展，但随后我遇到了父/子类问题，因为您无法从 OneToMany/ManyToOne 引用中引用 MappedSuperclass。

所以，终于进入正题了:有没有办法以更好/更干净的方式实现这一点？这些警告有意义还是只是 Intellij 不够聪明？

Answer 1

问题不在于 JPA，而在于泛型的使用。

首先，更改抽象类签名，使其具有递归类型:

public abstract class TreeStructure<T extends TreeStructure<T>>

接下来，您无法引用“this”，因为您不知道“this”的实现，因此您可以将其强制转换为“T”，或者添加带有如下签名的抽象方法:

public abstract T getImpl();

在实现中只返回“this”。

public T getImpl() {
  return this;
}

在侧节点上，访问类中的父类实例变量可能不是一个好主意。向 TreeStructure 类添加 addChild 和 removeChild 方法可能是一个更好的主意。