[英]Select query with JPA Criteria API When Entity class has a child class (Embeddable id)
[英]How to select a JPA Entity by whitelisting a property of its child Entity with the Criteria API?
我坚持以下假设问题:
使用Criteria API (而不是JPQL ),并给出
一个充满用户的桌子,每个用户都有多辆车
@Entity public class User { @Id private Long id; @OneToMany private Set<Car> cars; }
@Entity public class Car { @Id private Long id; private String model; }
包含汽车模型的 Set<String>
:
Set<String> models = getThousandsOfModels();
我怎样才能选择
User
小号具有至少一个Car
,其model
是在models
设置?
我已经阅读了许多SO答案,尝试了不同的方式(最有希望的似乎是使用Expression<Collection<String>>
创建谓词,使用.in()
等)但没有任何效果。
以下代码可以解决问题:
Set<String> models = getHundredsOfModels();
if (models.size()>1000) throw new TooManyResultsException(); // Oracle limit
CriteriaBuilder cb = em.getCriteriaBuilder();
CriteriaQuery<User> cQuery = cb.createQuery(User.class);
Root<User> user = cQuery.from(User.class);
List<Predicate> predicates = new ArrayList<Predicate>();
// This lines are the trick ------------------------------------------
Join<User, Car> usersCars = user.join("cars", JoinType.INNER);
Predicate p = usersCars.<String>get("model").in(models);
// ---------------------------------------------------------------------
predicates.add(p);
cQuery.select(user).where(predicates.toArray(new Predicate[predicates.size()]));
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