Spring CrudRepository save excluding child entity in JSON? - json

Question

I have a SpringBoot app and am using the CrudRepository to persist objects to the DB. I have an Product entity class which has a many to one relationship with a Vendor entity. I'm passing JSON that includes the details of both Product and the embedded vendor, but I'd ultimately like to pass just the Product details and include the vendorID. Is there some annotation that can resolve this for me?
Here is my code:
#Entity
public class Product {
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
private int id;
#ManyToOne
#JoinColumn(name="VendorID")
private Vendor vendor;
#Column(name="partnumber")
#JsonProperty("PartNumber")
private String vendorPartNumber;
#JsonProperty("Name")
private String name;
#Entity
public class Vendor {
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
private int id;
#JsonProperty("Code")
private String code;
#JsonProperty("Name")
private String name;
....
#OneToMany(mappedBy = "vendor", cascade = CascadeType.ALL, fetch = FetchType.EAGER)
#JsonIgnore
private List<Product> products;
And the JSON I'm passing (which works) looks like this:
{
"Vendor": {
"Id":1,
"Code": "BB-1001",
"Name": "Best Buy",
"Address": "100 Best Buy Street",
"City": "Louisville",
"State": "KY",
"Zip": "40207",
"Phone": "502-111-9099",
"Email": "geeksquad#bestbuy.com",
"IsPreApproved": "false"
},
"PartNumber":"TEST01",
"Name":"Test Product 01",
"Price":99.99
}
I'd ultimately like to remove the JSON object reference to vendor and replace with VendorId.

Answer 1

Solution by annotation
You can use #JsonSerialize to provide a custom serialization to achieve what you want.
#Entity
public class Product {
...
#JsonSerialize(using = VendorToIdSerializer.class)
#ManyToOne
#JoinColumn(name="VendorID")
private Vendor vendor;
...
}
public class VendorToIdSerializer extends JsonSerializer<Vendor> {
#Override
public void serialize(Vendor vendor, JsonGenerator jsonGenerator, SerializerProvider serializerProvider) throws IOException, JsonProcessingException {
String vendorIdAsString = String.valueOf(vendor.getId());
jsonGenerator.writeObject(vendorIdAsString);
}
}
Solution by DTO
However, as others have pointed out, the cleaner approach would be to create a DTO class which serves the role of the "view" of Product.
public class ProductView {
private final int vendor;
...
public ProductView(Vendor vendor) {
this.vendor = vendor.getId();
...
}
// getters
}

Answer 2

My issue was answered by using the #JsonBackReference and #JsonManagedReference annotations. I'm going to continue to research the DTOs as suggested above though. Thank-you!

Answer 3

I would also suggest to create a DTO that fit your needs.
With JPA you can create DTO instances using the constructor expression:
Here an example from a great article written by Vlad Mihalcea
https://vladmihalcea.com/the-best-way-to-map-a-projection-query-to-a-dto-with-jpa-and-hibernate/
List<PostDTO> postDTOs = entityManager.createQuery(
"select new " +
" com.vladmihalcea.book.hpjp.hibernate.query.dto.projection.jpa.PostDTO(" +
" p.id, " +
" p.title " +
" ) " +
"from Post p " +
"where p.createdOn > :fromTimestamp", PostDTO.class)
.setParameter( "fromTimestamp", Timestamp.from(LocalDateTime.of( 2016, 1, 1, 0, 0, 0 )
.toInstant( ZoneOffset.UTC ) ))
.getResultList();