如何使用JPA Criteria Builder编写查询(包括子查询和存在)

时间:2013-10-07 10:49:44

标签: hibernate jpa jpa-2.0 criteria

努力使用JPA编写以下查询。

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');

另外,您是否希望指出任何优秀的教程来编写JPA中的复杂查询。

2 个答案:

答案 0 :(得分:6)

我将使用 JpaRepository,JpaSpecificationExecutor,CriteriaQuery,CriteriaBuilder 回答简单汽车广告域(广告,品牌,型号)的示例:

  • 品牌[one-to-many]型号
  • 模型[一对多]广告

<强>实体:

@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;
}

<强>存储库:

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

<强>规格:

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);
  }
}

效果就是这个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=?)

答案 1 :(得分:1)

使用JPA Queries或HQL代替Criteria构建器可以更简单地执行此操作:

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')