java - JPA Criteria 根据条件过滤集合

Question

为什么 Criteria 查询不基于服务(集合实体)条件services.get("status"), "pending")过滤记录，如下所示？

CriteriaQuery<Customer> query = cb.createQuery(Customer.class);
    Root<Customer> customer = query.from(Customer.class);
Join<Customer, Service> services = customer.join("services", JoinType.INNER);

List<Predicate> predicates = new ArrayList<Predicate>();

predicates.add(cb.equal(customer.get("customerId"), 1));
predicates.add(cb.equal(services.get("status"), "pending"));

query.select(customer).distinct(true)
        .where(predicates.toArray(new Predicate[]{}));

List<Customer> customers = em.createQuery(query).getResultList();

SQL 正确过滤记录的位置

  select * from customers c
  INNER JOIN SERVICES s on s.COID = c.COID 
  where c.ID=1 and
  s.status='pending';

根据状态条件(用于收集)，结果集中的记录不合格，实际上，客户的所有服务都被退回。

我尝试使用 fetch Join (因为第一个为客户执行了 2 个查询，第二个为该客户的服务执行了第二个查询，认为第二个查询中可能不会评估条件)使用

customer.fetch("services", JoinType.INNER);

但运气不好。

我对这种行为感到惊讶。我正在使用 OpenJPA JPA 提供程序

实体是客户和服务。

 public class Customer{
    @Id
    @Column(name = "ID")
    private Integer customerId;
    @OneToMany
    @MappedBy(name = "customer")
    private List<Service> services;
    }

 public class Service {
    @EmbeddedId
    private ServicesPK servicePK;
    @ManyToOne
    @JoinColumn(name = "COID")
    private Customer customer;
    }

 @Embeddable
 @EqualsAndHashCode
 public class ServicesPK implements Serializable {
    @Column(name = "COID")
    private Integer coId;
    @Column(name = "VERSION")
    private Integer version;
}

Answer 1

尝试这段代码，我所做的更改是我向查询添加了类型安全。了解类型安全 typesafe

CriteriaQuery<Customer> query = cb.createQuery(Customer.class);
Root<Customer> customer = query.from(Customer.class);
Join<Customer, Service> services = customer.join(Customer_.services);
List<Predicate> predicates = new ArrayList<Predicate>();

predicates.add(cb.equal(customer.get("customerId"), 1));
predicates.add(cb.equal(services.get(Service_.status), "pending"));

query.select(customer).distinct(true)
    .where(predicates.toArray(new Predicate[]{}));

List<Customer> customers = em.createQuery(query).getResultList();