java 创建ManyToMany关系的两个表中的ManyToMany组合键如何对其进行注解

6l7fqoea  于 2023-02-18  发布在  Java
关注(0)|答案(1)|浏览(126)

bounty将在6天后过期。回答此问题可获得+50声望奖励。BugsOverflow希望引起更多人对此问题的关注:期待一个答案,显示我拥有的模型类应该是什么样子的,应该基于我描述的多对多关系

我有一个多对多的关系之间的树和节点:

  • 树可以有许多节点。
  • 节点可以位于多个树中。

在Tree表和Node表中也有复合主键。
所以我的模型类看起来像这样:

@Entity
@Table(name = "tree")
@AllArgsConstructor
@NoArgsConstructor
@Getter
@Setter
@JsonNaming(PropertyNamingStrategies.SnakeCaseStrategy.class)
public class Tree {
    @EmbeddedId
    private TreeIdentifier treeIdentifier;
    private LocalDateTime creationDate;
}

树标识符(这是树的主键):

@Embeddable
@AllArgsConstructor
@NoArgsConstructor
@Getter
@Setter
@JsonNaming(PropertyNamingStrategies.SnakeCaseStrategy.class)
public class TreeIdentifier implements Serializable {

    private String treeId;
    private String siteId;
    private Integer prodVersion;

    @Override
    public int hashCode() {
        return Objects.hash(prodVersion, siteId, treeId);
    }

    @Override
    public boolean equals(Object obj) {
        if (this == obj)
            return true;
        if (obj == null)
            return false;
        if (getClass() != obj.getClass())
            return false;
        TreeIdentifier other = (TreeIdentifier) obj;
        return Objects.equals(prodVersion, other.prodVersion) && Objects.equals(siteId, other.siteId)
                && Objects.equals(treeId, other.treeId);
    }

    private static final long serialVersionUID = 1L;
}

节点:

@Entity
@Table(name = "node")
@AllArgsConstructor
@NoArgsConstructor
@Getter
@Setter
@JsonNaming(PropertyNamingStrategies.SnakeCaseStrategy.class) 
public class Node {
    @EmbeddedId
    private NodeIdentifier nodeIdentifier;
    private Long nodePosition;
}

节点标识符(节点的主键):

@Embeddable
@AllArgsConstructor
@NoArgsConstructor
@Getter
@Setter
@JsonNaming(PropertyNamingStrategies.SnakeCaseStrategy.class)
public class NodeIdentifier implements Serializable {

    private String nodeId;
    private Integer nodeVersionId;

    @Override
    public int hashCode() {
        return Objects.hash(nodeId, nodeVersionId);
    }

    @Override
    public boolean equals(Object obj) {
        if (this == obj)
            return true;
        if (obj == null)
            return false;
        if (getClass() != obj.getClass())
            return false;
        NodeIdentifier other = (NodeIdentifier) obj;
        return Objects.equals(nodeId, other.nodeId) && Objects.equals(nodeVersionId, other.nodeVersionId);
    }

    private static final long serialVersionUID = 1L;

}

现在,我正在努力弄清楚由于这个ManyToMany关系而激增的中间表在代码中应该是什么样子的:

树节点

@Entity
@Table(name = "tree_node")
@AllArgsConstructor
@NoArgsConstructor
@Getter
@Setter
@JsonNaming(PropertyNamingStrategies.SnakeCaseStrategy.class)
public class TreeNode {
    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private Integer treeNodeId;
    private Tree tree;
    private Node node;
}

我是按照Baeldung指南,到目前为止,我设法使它的工作与保存树和节点在数据库中,但现在的中间表不能弄清楚我应该如何注解字段正确(也应该如何注解树和节点类太?),你能帮助我吗?
我使用的指南:

n3schb8v

n3schb8v1#

您可以使用@MapsId;也许这就是你想要达到的目标。

public class TreeNode {
    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private Integer treeNodeId;
    @ManyToOne
    @MapsId("treeIdentifier")
    private Tree tree;
    @ManyToOne
    @MapsId("nodeIdentifier")
    private Node node;

}

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