java - JPA @onetomany 级联插入正在抛出 org.hibernate.exception.ConstraintViolationException

Question

附属类:

@Entity
@Table(name="attachments")
@Getter
@Setter
public class AttachmentModel {

    //@EmbeddedId
    //private AttachmentId attachmentId;

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    @Column(name="notice_attachment_id")    
    private long attachmentId;

    @Column(name="notice_id")
    private long noticeId;

    @Column(name="attachment")
    private String attachmentUrl;

    @JsonIgnore
    @ManyToOne(cascade = {CascadeType.PERSIST , CascadeType.MERGE,
            CascadeType.DETACH , CascadeType.REFRESH},optional = false)
    @JoinColumn(name="notice_id", insertable=false, updatable=false)
    @MapsId("notice_id")
    NoticesModel notice;

    public void addNotice(NoticesModel notice) {
        this.notice = notice;
    }

    public AttachmentModel() {

    }   
}

通知类:

@Entity
@Table(name = "notices")
@Getter @Setter
public class NoticesModel {

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    @Column(name = "notice_id" ,updatable = false, nullable = false,insertable = true)
    private long noticeID;

    @OneToMany(fetch = FetchType.EAGER, cascade = { CascadeType.ALL } , mappedBy = "notice")
    //@mappedBy(name = "notice_id")
    private List<AttachmentModel> attachments;
}

解析JSON并保存的代码

public HashMap<String,Object> saveNotices(@RequestBody List<NoticesModel> tmpNotices)
    {
        List<NoticesModel> notices = tmpNotices;
        for (NoticesModel notice : notices) {
            List<AttachmentModel> attachments =  notice.getAttachments();
            for (AttachmentModel attachment : attachments) {
                attachment.addNotice(notice);
                System.out.println(attachment.getAttachmentUrl());
            }

            for (AttachmentModel attachment : attachments) {
                //attachment.addNotice(notice);
                System.out.println(attachment.getNotice().getContent());
                System.out.println(attachment.getNotice().getNoticeID());
            }
        }
        int result = noticesServices.saveNotice(notices);

        HashMap<String,Object> res = new HashMap<>();
        res.put("message",result);
        return res;

    }

这是我要发送的 JSON

[
  {
    "attachments": [
      {
        "attachmentUrl": "/abc/bcd"
      }
    ],
    "content": "string",
  }
]

对于这种情况，我试图保存我的通知和附件。在这种特殊情况下，notice_id 是在保存到数据库时创建的。

因此在尝试保存附件表时，它试图将 notice_id 保存为 0。

所以我得到了异常(exception)。

无法执行语句； SQL [不适用]；约束 [attachments_notices_fk]；嵌套异常是 org.hibernate.exception.ConstraintViolationException: 无法执行语句

我怎样才能解决这个问题？是否可以在保存到数据库之前获取 notice_id，以便我可以获得 notice_id 以便我可以将其设置在附件中，这样它就不会被保存为 0？在这种情况下(我是 JPA 和 springboot 的新手)我做错了什么(我可以采取任何替代方法)？

Answer 1

我认为您不需要使用任何 notice_id。从你的 AttachmentModel 中删除 notice_id 和相关的东西，并使用 notice 进行映射(注意:仍然会有列 notice_id在删除后的数据库中)，所以:

@ManyToOne
private NoticesModel notice;

并更改 NoticesModel 中的映射以引用正确的字段:

//                                    ALL is just a suggestion
@OneToMany(mappedBy = "noticesModel", cascade = CascadeType.ALL)
private List<AttachmentModel> attachementModels;

那么你的 for 循环可能看起来像:

for (NoticesModel notice : notices) {
    for (AttachmentModel am : notice.getAttachments()) {
        am.setNotice(notice);
    }
    noticesServices.save(notice);
}

您还可以在 NoticesModel 中添加类似这样的内容，以便在持久化之前始终处理设置引用:

@PrePersist
private void prePersist() {
    for (AttachmentModel am : attachments) {
        am.setNotice(this);
    }
}