How to write query(include subquery and exists) using JPA Criteria Builder

问题

Struggling to write the following query using JPA.

Oracle Query:

Select * from table1 s
where exists (Select 1 from table2 p
              INNER JOIN table3 a ON a.table2_id = p.id
              WHERE a.id = s.table3_id
              AND p.name = 'Test');

Also, would you like to point any good tutorial to write complex queries in JPA.

回答1:

I will answer the example of the simple car advertisement domain (advert, brand, model) using JpaRepository, JpaSpecificationExecutor, CriteriaQuery, CriteriaBuilder:

brand [one-to-many] model
model [one-to-many] advert

Entities:

@Entity
public class Brand {
  @Id
  @GeneratedValue(strategy = GenerationType.AUTO)
  private Long id;
  private String name;
  @OneToMany(mappedBy = "brand", fetch = FetchType.EAGER)
  private List<Model> models;
}

@Entity
public class Model {
  @Id
  @GeneratedValue(strategy = GenerationType.AUTO)
  private Long id;
  private String name;
  @ManyToOne
  @JoinColumn(name = "brand_id")
  private Brand brand;
}

@Entity
public class Advert {
  @Id
  @GeneratedValue(strategy = GenerationType.AUTO)
  private Long id;
  @ManyToOne
  @JoinColumn(name = "model_id")
  private Model model;
  private int year;
  private int price;
}

Repository:

public interface AdvertRepository
  extends JpaRepository<Advert, Long>, JpaSpecificationExecutor<Advert> {
}

Specification:

public class AdvertSpecification implements Specification<Advert> {
  private Long brandId;

  public AdvertSpecification(Long brandId) {
    this.brandId = brandId;
  }

  @Override
  public Predicate toPredicate(Root<Advert> root,
                               CriteriaQuery<?> query,
                               CriteriaBuilder builder) {

    Subquery<Model> subQuery = query.subquery(Model.class);
    Root<Model> subRoot = subQuery.from(Model.class);

    Predicate modelPredicate = builder.equal(root.get("model"), subRoot.get("id"));

    Brand brand = new Brand();
    brand.setId(brandId);
    Predicate brandPredicate = builder.equal(subRoot.get("brand"), brand);

    subQuery.select(subRoot).where(modelPredicate, brandPredicate);
    return builder.exists(subQuery);
  }
}

Effect is this Hibernate SQL:

select advert0_.id as id1_0_,
       advert0_.model_id as model_id5_0_,
       advert0_.price as price3_0_,
       advert0_.year as year4_0_
from advert advert0_
where exists (select model1_.id from model model1_
              where advert0_.model_id=model1_.id
              and model1_.brand_id=?)

回答2:

You can do it much simpler using JPA Queries or HQL instead of Criteria builders:

SELECT e1 from Entity1 as e1 
where exists
(select e2 from Entity2 as e2 join e2.e3 as ent3
where ent3.id=e1.id and e2.name='Test')

来源：https://stackoverflow.com/questions/19222868/how-to-write-queryinclude-subquery-and-exists-using-jpa-criteria-builder

标签

hibernate

jpa

jpa-2.0

criteria