使用 JPA 创建树

Question

我正在尝试使用实体和 JPA 创建一棵树。我有一个具有以下属性的类。

public class Dir
{

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

@OneToOne(mappedBy="dirId", cascade= CascadeType.ALL)
private Dir parent;
...

一个节点知道它的父节点是谁，如果它没有父节点，那么它就是根。这就是我如何轻松地用它构建一棵树的方法。但是......我认为这个想法的映射是不正确的。当我尝试部署时出现以下错误：

An incompatible mapping has been encountered between [class com.dv.oa.model.entity.dir.Dir] and [class com.dv.oa.model.entity.dir.Dir]. This usually occurs when the cardinality of a mapping does not correspond with the cardinality of its backpointer.

它谈论基数。但这没有意义，一个节点只能有一个父节点。这就是我选择的原因@OneToOne

有人能解释一下吗？我想问这个问题的另一种方式是，如何将一个实体映射到其自身的另一个实例？

编辑

这是我的表结构：

mysql> describe dir;
+--------------+--------------+------+-----+---------+-------+
| Field        | Type         | Null | Key | Default | Extra |
+--------------+--------------+------+-----+---------+-------+
| dirId        | int(11)      | NO   | PRI | NULL    |       |
| DTYPE        | varchar(31)  | YES  |     | NULL    |       |
| dirName      | varchar(255) | YES  |     | NULL    |       |
| companyOwner | int(11)      | YES  | MUL | NULL    |       |
| userOwner    | int(11)      | YES  | MUL | NULL    |       |
| parent       | int(11)      | YES  |     | NULL    |       |
+--------------+--------------+------+-----+---------+-------+
6 rows in set (0.00 sec)

Answer 1

您指向映射所属方的错误列。此外，你们的关系不是一对一的，因为单亲可以有很多孩子。

@Entity
public class Dir
{

  //This field is a table column
  //It uniquely identifies a row on the DIR table
  @Id
  private int dirId;

  //This field is a table column
  // It identifies the parent of the current row
  // It it will be written as the type of dirId
  // By default this relationship will be eagerly fetched
  // , which you may or may not want
  @ManyToOne(fetch=FetchType.LAZY, cascade={CascadeType.PERSIST, CascadeType.MERGE})
  private Dir parent;

  //This field is not a table column
  // It is a collection of those Dir rows that have this row as a parent. 
  // This is the other side of the relationship defined by the parent field.
  @OneToMany(mappedBy="parent")
  private Set<Dir> children;
}