如何使用JPA Criteria Builder编写查询（包括子查询和存在查询） [英] How to write query(include subquery and exists) using JPA Criteria Builder

问题描述

正在努力使用JPA编写以下查询。

Struggling to write the following query using JPA.

Oracle查询：

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');

此外，您想指出任何好的教程来用JPA编写复杂的查询。

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

推荐答案

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

• 品牌[一对多]模型


• 模型[一个-to-many]广告

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;
}

存储库：

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

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);
  }
}

效果是此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=?)

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