Question

无论如何使用提交的注释为项目字段设置ID并获取项目对象。所以我只能设置项目ID ，同时坚持这个子表而不是设置整个对象，因为我们要在board_project表中只保存项目的id。我是一个副本，我找不到其他链接。这就像我们正在使用提交注释而不是方法。

使用：hibernate5.0 - jpa2.1 lombok（即getter / setter）for Spring-data-commons-1.13 for CRUD Operations

import lombok.AllArgsConstructor;
import lombok.Getter;
import lombok.NoArgsConstructor;
import lombok.Setter;

import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.GeneratedValue;
import javax.persistence.Id;
import javax.persistence.JoinColumn;
import javax.persistence.ManyToOne;
import javax.persistence.SequenceGenerator;
import javax.persistence.Table;

@Getter
@Setter
@NoArgsConstructor
@AllArgsConstructor
@Entity
@Table(name = "board_project")
public class BoardProject {

    @Id
    @SequenceGenerator(name = "board_project_id_generator", sequenceName = "board_project_id_seq", allocationSize = 1)
    @GeneratedValue(generator = "board_project_id_generator")
    private Long id;

    @Column(name = "board_id")
    private Long boardId;

    @ManyToOne
    @JoinColumn(name = "project_id")
    private Project project;          // field set id and get object
   }

项目实体：

public class Project{

    @Id
    @SequenceGenerator(name = "project_id_generator", sequenceName = "project_id_seq", allocationSize = 1)
    @GeneratedValue(generator = "project_id_generator")
    private Long id;

    @Column(name = "name")
    private String name;
}

更新：发现类似这样的内容

@ManyToOne
@JoinColumn(name = "project_id", updatable = false, insertable = false)
private Project project;

@Column(name = "project_id")
@NotNull
private Long projectId;

不知道是否可以

对此的任何帮助都会很棒。

Hibernate Field注释joinColumn设置id和get对象

0 个答案: