i want to store the JSON using spring boot with JPA
{ "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
spring spring-boot spring-data-jpa
|
show 8 more comments
{ "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
spring spring-boot spring-data-jpa
2
What have you done yet?
– Alex Foglia
Nov 12 at 5:37
i have created the data model and then through save method i am calling
– Shashank Ranjan
Nov 12 at 5:46
controller--> service--->Repo <myEntiyclass, Integer>
– Shashank Ranjan
Nov 12 at 5:47
please share your code
– danny
Nov 12 at 5:48
controller class @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); }
– Shashank Ranjan
Nov 12 at 5:55
|
show 8 more comments
{ "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
spring spring-boot spring-data-jpa
{ "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
spring spring-boot spring-data-jpa
spring spring-boot spring-data-jpa
edited Nov 12 at 13:38
BalusC
840k29631153194
840k29631153194
asked Nov 12 at 5:32
Shashank Ranjan
25
25
2
What have you done yet?
– Alex Foglia
Nov 12 at 5:37
i have created the data model and then through save method i am calling
– Shashank Ranjan
Nov 12 at 5:46
controller--> service--->Repo <myEntiyclass, Integer>
– Shashank Ranjan
Nov 12 at 5:47
please share your code
– danny
Nov 12 at 5:48
controller class @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); }
– Shashank Ranjan
Nov 12 at 5:55
|
show 8 more comments
2
What have you done yet?
– Alex Foglia
Nov 12 at 5:37
i have created the data model and then through save method i am calling
– Shashank Ranjan
Nov 12 at 5:46
controller--> service--->Repo <myEntiyclass, Integer>
– Shashank Ranjan
Nov 12 at 5:47
please share your code
– danny
Nov 12 at 5:48
controller class @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); }
– Shashank Ranjan
Nov 12 at 5:55
2
2
What have you done yet?
– Alex Foglia
Nov 12 at 5:37
What have you done yet?
– Alex Foglia
Nov 12 at 5:37
i have created the data model and then through save method i am calling
– Shashank Ranjan
Nov 12 at 5:46
i have created the data model and then through save method i am calling
– Shashank Ranjan
Nov 12 at 5:46
controller--> service--->Repo <myEntiyclass, Integer>
– Shashank Ranjan
Nov 12 at 5:47
controller--> service--->Repo <myEntiyclass, Integer>
– Shashank Ranjan
Nov 12 at 5:47
please share your code
– danny
Nov 12 at 5:48
please share your code
– danny
Nov 12 at 5:48
controller class @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); }
– Shashank Ranjan
Nov 12 at 5:55
controller class @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); }
– Shashank Ranjan
Nov 12 at 5:55
|
show 8 more comments
1 Answer
1
active
oldest
votes
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 (?, ?, ?, ?, ?)
add a comment |
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1 Answer
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1 Answer
1
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active
oldest
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active
oldest
votes
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 (?, ?, ?, ?, ?)
add a comment |
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 (?, ?, ?, ?, ?)
add a comment |
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 (?, ?, ?, ?, ?)
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 (?, ?, ?, ?, ?)
answered Nov 12 at 6:20
vishal lakhyani
694
694
add a comment |
add a comment |
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2
What have you done yet?
– Alex Foglia
Nov 12 at 5:37
i have created the data model and then through save method i am calling
– Shashank Ranjan
Nov 12 at 5:46
controller--> service--->Repo <myEntiyclass, Integer>
– Shashank Ranjan
Nov 12 at 5:47
please share your code
– danny
Nov 12 at 5:48
controller class @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); }
– Shashank Ranjan
Nov 12 at 5:55