i want to store the JSON using spring boot with JPA - spring

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"
}
]
}
My controller class is
#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);
}
service class
public void addFavouriteData(FavouriteDTO requestInputMapper)
{
favouriteRepository.save(requestInputMapper);
}
And these are entity class !!
#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;
}
And
#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;
}
here the id is auto genrated. and the favouriteId is same as id.
so how can i store the data as id is auto genrated and i need to put the same favourite id as in id. so how can i store the data in the data base
so i have given my entity class. i have two entity Favorite and DefaultFavuorite Entity.so how can i store the data

Answer 1

You can tell Hibernate, and any other JPA implementation, to cascade certain operations you perform on an entity to its associated child entities. The only thing you have to do is to define the kind of operation you want to cascade to the child entities.
The following code snippet shows an example in which I cascade the persist operation of the Author entity to all associated Book entities.
#Entity
public class Author {
…
#ManyToMany(mappedBy=”authors”, cascade = CascadeType.PERSIST)
private List<Book> books = new ArrayList<Book>();
…
}
When you now create a new Author and several associated Book entities, you just have to persist the Author entity.
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();
As you can see in the log output, Hibernate cascades the operation to the associated Book entities and persists them as well.
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 (?, ?, ?, ?, ?)