Question

     {      "id" :"3",
            "userId": "abc",
            "favName": "shashank",
            "kpiName": "FavKPI",
            "rptId": "529",
            "language": "EN",
            "selectedControlIdList": [
                {

                    "favouriteId": 3,
                    "controlId": "3",
                    "controlName": " ",
                    "label": "Plant",
                    "keyValue": "KPI_01_PL_01_1",
                    "structureType": "LISTBOX"
                },
                {

                    "favouriteId": 3,
                    "controlId": "2",
                    "controlName": " ",
                    "label": "Plant12",
                    "keyValue": "KPI_01",
                    "structureType": "LISTBOX"
                }
            ]
        }

我的控制器类是

    @RequestMapping(value = "/addFavData", method = RequestMethod.POST, consumes =MediaType.APPLICATION_JSON_VALUE, produces =MediaType.APPLICATION_JSON_VALUE) 
        public void addFavData(@RequestBody FavouriteDTO requestInputMapper) { 
          favouriteService.addFavouriteData(requestInputMapper); 
    }

服务等级

     public void addFavouriteData(FavouriteDTO requestInputMapper) 
        { 
          favouriteRepository.save(requestInputMapper); 
        }

这些是实体类！

@Entity
@Table(name = "favorite", schema = "quality")
public class FavouriteDTO implements Serializable{


    private static final long serialVersionUID = -7089417397407966229L;
    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private Integer id;
    @Column(name = "userId")
    private String userId;
    @Column(name = "favName")
    private String favName;
    @Column(name = "kpiName")
    private String kpiName;
    @Column(name = "rptId")
    private String rptId;
    @Column(name = "language")
    private String language;

    @OneToMany(cascade = CascadeType.ALL, fetch = FetchType.EAGER)
    @JoinColumn(name = "favouriteId")
    private List<DefaultControlsDTO> selectedControlIdList;
}

和

@Entity
@Table(name = "favoriteControls", schema = "quality")
public class DefaultControlsDTO implements Serializable {


    private static final long serialVersionUID = 8720721227933753311L;
    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private Integer id;

    @Column(name = "favouriteId")
    private Integer favouriteId;

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

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

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

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

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

}

此处ID是自动生成的。并且favouriteId与id相同。所以我该如何存储数据，因为id是自动生成的，因此我需要输入与id相同的收藏ID。所以我如何将数据存储在数据库中所以我给了我的实体课。我有两个实体收藏夹和DefaultFavuorite实体。所以我该如何存储数据

Answer 1

您可以告诉Hibernate和任何其他JPA实现，将您在实体上执行的某些操作级联到与其关联的子实体。唯一要做的就是定义要级联到子实体的操作类型。

下面的代码片段显示了一个示例，其中我将Author实体的持久操作层叠到所有关联的Book实体。

    @Entity
    public class Author {

        …

        @ManyToMany(mappedBy=”authors”, cascade = CascadeType.PERSIST)
        private List<Book> books = new ArrayList<Book>();

        …
    }

现在创建新的Author和几个关联的Book实体时，只需保留Author实体。

EntityManager em = emf.createEntityManager();
em.getTransaction().begin();

Author a = new Author();
a.setFirstName(“John”);
a.setLastName(“Doe”);

Book b1 = new Book();
b1.setTitle(“John’s first book”);
a.getBooks().add(b1);

Book b2 = new Book();
b2.setTitle(“John’s second book”);
a.getBooks().add(b2);

em.persist(a);

em.getTransaction().commit();
em.close();

正如您在日志输出中看到的那样，Hibernate将操作级联到关联的Book实体，并将其持久化。

15:44:28,140 DEBUG [org.hibernate.SQL] – insert into Author (firstName, lastName, version, id) values (?, ?, ?, ?)
15:44:28,147 DEBUG [org.hibernate.SQL] – insert into Book (publisherid, publishingDate, title, version, id) values (?, ?, ?, ?, ?)
15:44:28,150 DEBUG [org.hibernate.SQL] – insert into Book (publisherid, publishingDate, title, version, id) values (?, ?, ?, ?, ?)

我想使用带有JPA的Spring Boot存储JSON

1 个答案: