CODEWITHSUNDEEP
jsonspringresthateoashal-json
Charlires
Charlires

Reputation: 873

Save related entities spring data rest HAL-JSON

I have the follow problem

I have a basic configuration of spring data rest (Nothing fancy, nothing custom).

Using spring-data-rest-webmvc 2.0.0 RELEASE and spring-data-jpa 1.5.0 RELEASE

Class A

@Entity
public class A {

   @Id
   @GeneratedValue(strategy = GenerationType.IDENTITY)
   private int id

   private String name;

   @ManyToMany
   private List<B> b;

   // getters setters
}

Class B

@Entity
public class B {

   @Id
   @GeneratedValue(strategy = GenerationType.IDENTITY)
   private int id

   private String nameb;

   @ManyToMany(mappedBy = "b")
   private List<A> a;

   // getters setters
}

Repository A

@Repository
@RestResource(rel = "a", path = "a")
public interface ARepository extends PagingAndSortingRepository<A, Integer> {

}

Repository B

@Repository
@RestResource(rel = "b", path = "b")
public interface BRepository extends PagingAndSortingRepository<B, Integer> {

}

When I save an entity works fine, but I don't know how to save a relationship

e.g. save an "A" inside a "B" using http

This is the last thing I try from this answer https://stackoverflow.com/a/13031580/651948

POST http://localhost:8080/api/a

{
    "name": "Name of A",
    "b": {
        "rel": "b",
        "href": "http://localhost:8080/api/b/1"
    }
}

I get an 201 http code but doesn't save the entity.

Did someone tried this already?

Upvotes: 4

Views: 1421

Answers (1)

JoeGo
JoeGo

Reputation: 293

Try just using the URL.

POST http://localhost:8080/api/a
Content-Type: application/json

{
    "name" : "Name of A",
    "b": "http://localhost:8080/api/b/1"
}

or, in your case, it's probably

"b" :  ["http://localhost:8080/api/b/1"]

because A.b is a list and hence you submit an array. Did not test this, though.

This should be the valid way since Spring 2.0 (see Spring Data Rest 2.0.0.RELEASE Breaks Code Working Previously With RC1) and it works for me well.

Upvotes: 1

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