Spring Data Join със спецификации

РЕДАКТИРАНЕ:Добре, доста се обърках тук, но се надявам този път да съм по-близо до правилния отговор.

Помислете (идентификаторите се генерират автоматично като 1 за Джон и т.н.):

INSERT INTO some_user (name) VALUES ('John');
INSERT INTO some_user (name) VALUES ('Ariel');
INSERT INTO some_user (name) VALUES ('Brian');
INSERT INTO some_user (name) VALUES ('Kelly');
INSERT INTO some_user (name) VALUES ('Tom');
INSERT INTO some_user (name) VALUES ('Sonya');

INSERT INTO product (owner_id,name) VALUES (1,'Nokia 3310');
INSERT INTO product (owner_id,name) VALUES (2,'Sony Xperia Aqua');
INSERT INTO product (owner_id,name) VALUES (3,'IPhone 4S');
INSERT INTO product (owner_id,name) VALUES (1,'Xiaomi MI5');
INSERT INTO product (owner_id,name) VALUES (3,'Samsung Galaxy S7');
INSERT INTO product (owner_id,name) VALUES (3,'Sony Xperia Z3');

INSERT INTO following_relationship (follower_id, owner_id) VALUES (4,1);
INSERT INTO following_relationship (follower_id, owner_id) VALUES (5,1);
INSERT INTO following_relationship (follower_id, owner_id) VALUES (4,2);
INSERT INTO following_relationship (follower_id, owner_id) VALUES (6,2);
INSERT INTO following_relationship (follower_id, owner_id) VALUES (6,3);
INSERT INTO following_relationship (follower_id, owner_id) VALUES (1,3);

Въз основа на опростена версия на обекти, които сте предоставили, и SomeUser Entity като:

@Entity
public class FollowingRelationship {

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

@ManyToOne(cascade=CascadeType.ALL, fetch=FetchType.EAGER)
@JoinColumn(name = "owner_id")
SomeUser owner;
    
@ManyToOne(cascade=CascadeType.ALL, fetch=FetchType.EAGER)
@JoinColumn(name = "follower_id")
SomeUser follower;

...

@Entity
public class Product {

@Id
@GeneratedValue(strategy=GenerationType.IDENTITY)
private Long id;
    
@ManyToOne()
@JoinColumn(name = "owner_id")
private SomeUser owner;
    
@Column
private String name;

...

@Entity
public class SomeUser {

@Id
@GeneratedValue(strategy=GenerationType.IDENTITY)
private Long id;
    
@Column
private String name;

@OneToMany(mappedBy = "owner")
private Set<Product> products = new HashSet<Product>();

@OneToMany(mappedBy = "owner")
private Set<FollowingRelationship> ownedRelationships = new HashSet<FollowingRelationship>();

@OneToMany(mappedBy = "follower")
private Set<FollowingRelationship> followedRelationships = new HashSet<FollowingRelationship>();

Създадох спецификация като:

public static Specification<Product> joinTest(SomeUser input) {
    return new Specification<Product>() {
        public Predicate toPredicate(Root<Product> root, CriteriaQuery<?> query, CriteriaBuilder cb) {
            Join<Product,SomeUser> userProd = root.join("owner");
            Join<FollowingRelationship,Product> prodRelation = userProd.join("ownedRelationships");
            return cb.equal(prodRelation.get("follower"), input);
        }
    };
}

И сега можем да изпълним заявката с:

SomeUser someUser = someUserRepository.findOne(Specification.where(ProductSpecifications.userHasName("Kelly")));
List<Product> thatProducts = productRepository.findAll(Specification.where(ProductSpecifications.joinTest(someUser)));
System.out.println(thatProducts.toString());

Получаваме:

[Product [id=1, name=Nokia 3310], Product [id=4, name=Xiaomi MI5], Product [id=2, name=Sony Xperia Aqua]]

И това според мен е еквивалентно на:"вземете всички продукти от всички потребители, които следва друг потребител" - вземете продукти на всички потребители, които Кели следва.