Fetch lazy object while serializing in spring data rest

Question

I have one entity A which has some properties, and reference to another entity B

A{
....
   @ManyToOne(fetch=FetchType.LAZY)
   B b;
...
}

same for B(@onetomany for A in B). Its a bidirectional relation. While calling the web service for entity A it gives link to the B entity instead of whole object in response json. I want the whole object in web service response JSON instead of link to that related entity. How can i do that? I am using springboot + spring data jpa + hibernate as jpa provider + spring 4

@dit : spring data rest provides url(link) for the nested objects . in this case it will provide url for B instead of object in json. — Kedar Parikh
– Kedar Parikh, Commented Mar 18, 2015 at 9:57
@dit : i changed to eager, but still B is not included in the A's web service response. — Kedar Parikh
– Kedar Parikh, Commented Mar 18, 2015 at 10:28

nKognito · Accepted Answer · 2015-06-05 07:22:58Z

In order to have nested related objects in the output JSON you have to use projections.

public class User {
   private String name;
   private Address address;
}

public class Address {
   private String street;
}

public interface UsersRepository extends JpaRepository<User, Long> {
   // Custom methods
}

will give you

{
   "name": "John",
   "_links": {
      "self": "http://localhost:8080/users/1",
      "address": "http://localhost:8080/addresses/1"
   }
}

BUT with projection:

@Projection(name = "inlineAddress", types = { User.class })
public interface UserInlineAddressProjection {
   String getName();
   Address getAddress();
}

@RepositoryRestResource(excerptProjection = UserInlineAddressProjection.class)
public interface UsersRepository extends JpaRepository<User, Long> {
   // Custom methods
}

result will be the following:

{
    "name": "John",
    "address": {
       "street": "Stormtroopers str. 1"
    },
    "_links": {
       "self": "http://localhost:8080/users/1",
       "address": "http://localhost:8080/addresses/1"
    }
}

This is the only way how you can implement it

habelson · Accepted Answer · 2015-08-15 03:39:38Z

0

It is unclear from your question that you are using a Collection on one side of the relationship (the "Many" in OneToMany).

There are two ways you can achieve this:

1) If you do not define a Repository for the child type, a link will not be generated, and it will instead be embedded in the parent object. Of course, this only is useful if you don't need a Repository for that object.

2) FetchType.EAGER will eagerly load the child object. Since the object is already loaded, there is no reason to provide a link to it, so spring-data-rest will embed the object. Make sure to put this on the side with the single object, not the Collection

answered Aug 15, 2015 at 3:39

habelson

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