i want to store the JSON using spring boot with JPA












0














     {      "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










share|improve this question




















  • 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


















0














     {      "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










share|improve this question




















  • 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
















0












0








0







     {      "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










share|improve this 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







spring spring-boot spring-data-jpa






share|improve this question















share|improve this question













share|improve this question




share|improve this question








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
















  • 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














1 Answer
1






active

oldest

votes


















0














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 (?, ?, ?, ?, ?)





share|improve this answer





















    Your Answer






    StackExchange.ifUsing("editor", function () {
    StackExchange.using("externalEditor", function () {
    StackExchange.using("snippets", function () {
    StackExchange.snippets.init();
    });
    });
    }, "code-snippets");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "1"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53256395%2fi-want-to-store-the-json-using-spring-boot-with-jpa%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0














    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 (?, ?, ?, ?, ?)





    share|improve this answer


























      0














      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 (?, ?, ?, ?, ?)





      share|improve this answer
























        0












        0








        0






        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 (?, ?, ?, ?, ?)





        share|improve this answer












        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 (?, ?, ?, ?, ?)






        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Nov 12 at 6:20









        vishal lakhyani

        694




        694






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Stack Overflow!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            To learn more, see our tips on writing great answers.





            Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


            Please pay close attention to the following guidance:


            • Please be sure to answer the question. Provide details and share your research!

            But avoid



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53256395%2fi-want-to-store-the-json-using-spring-boot-with-jpa%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            Popular posts from this blog

            Florida Star v. B. J. F.

            Danny Elfman

            Retrieve a Users Dashboard in Tumblr with R and TumblR. Oauth Issues