JPA - Criteria API 和 EmbeddedId

Question

我想使用条件进行以下查询。我有一个定义了 EmbeddedId 的 Entity:

 @Entity
 @Table(name="TB_INTERFASES")
 public class Interfase implements Serializable {

  @EmbeddedId
  private InterfaseId id;
 }

 @Embeddable
 public class InterfaseId implements Serializable {
  @Column(name="CLASE")
  private String clase;
 }

我想做的条件查询是:

 CriteriaBuilder criteriaBuilder = this.entityManager.getCriteriaBuilder();
 CriteriaQuery<Interfase> criteriaQuery = criteriaBuilder.createQuery(Interfase.class);
 Root<Interfase> entity = criteriaQuery.from(Interfase.class);
 criteriaQuery.where(
   criteriaBuilder.equal(entity.get("clase"), "Clase"),
 );

但这会引发 IllegalArgumentException:

java.lang.IllegalArgumentException: Not an managed type: class InterfaseId

我也尝试过这个查询:

 Root<Interfase> entity = criteriaQuery.from(Interfase.class);
 criteriaQuery.where(
   criteriaBuilder.equal(entity.get("id").get("clase"), "Clase"),
 );

还有这个...

 Root<Interfase> entity = criteriaQuery.from(Interfase.class);
 criteriaQuery.where(
   criteriaBuilder.equal(entity.get("id.clase", "Clase"),
 );

没有运气。所以我的问题是，当我的类使用 Embedded 和 EmbeddedId 注释时，如何使用条件进行查询？

谢谢!毛罗。

Answer 1

您需要使用路径导航来访问Embeddable的属性。以下是 JPA 2.0 规范中的示例(使用静态元模型):

6.5.5 Path Navigation

...

In the following example, ContactInfo is an embeddable class consisting of an address and set of phones. Phone is an entity.

CriteriaQuery<Vendor> q = cb.createQuery(Vendor.class);
Root<Employee> emp = q.from(Employee.class);
Join<ContactInfo, Phone> phone =
    emp.join(Employee_.contactInfo).join(ContactInfo_.phones);
q.where(cb.equal(emp.get(Employee_.contactInfo)
                    .get(ContactInfo_.address)
                    .get(Address_.zipcode), "95054"))
    .select(phone.get(Phone_.vendor));

The following Java Persistence query language query is equivalent:

SELECT p.vendor
FROM Employee e JOIN e.contactInfo.phones p
WHERE e.contactInfo.address.zipcode = '95054'

所以就你的情况而言，我认为你需要这样的东西:

criteriaBuilder.equal(entity.get("id").get("clase"), "Referencia 111")

引用文献

• JPA 2.0 规范
  • 第 6.5.5 节“路径导航”

<小时/>

更新:我已经使用 Hibernate EntityManager 3.5.6 和以下查询测试了提供的实体:

CriteriaBuilder builder = em.getCriteriaBuilder();

CriteriaQuery<Interfase> criteria = builder.createQuery(Interfase.class);
Root<Interfase> interfaseRoot = criteria.from(Interfase.class);
criteria.select(interfaseRoot);
criteria.where(builder.equal(interfaseRoot.get("id").get("clase"), 
    "Referencia 111"));

List<Interfase> interfases = em.createQuery(criteria).getResultList();

运行良好并生成以下 SQL:

17:20:26.893 [main] DEBUG org.hibernate.SQL -     select        interfase0_.CLASE as CLASE31_     from        TB_INTERFASES interfase0_     where        interfase0_.CLASE=?17:20:26.895 [main] TRACE org.hibernate.type.StringType - binding 'Referencia 111' to parameter: 1

按预期工作。