如何在JPA 2.0中为多个键值组合子查询地图？

Question

我有以下实体（示例）：

@Entity
@Table(name = "person")
public class Person implements Serializable {

  @Id
  @Column(name = "person_id", columnDefinition = "UUID")
  private UUID userId;

  @Column(name = "name")
  private String name;


  @ElementCollection
  @MapKeyColumn(name = "phonetype")
  @Column(name = "number")
  @CollectionTable(name = "person_phones", joinColumns = @JoinColumn(name = "userId"))
  private Map<String, String> phoneNumbers;

}

现在，在这个例子中，phoneNumbers是String，String。我们假设密钥是类型（如“移动”，“主页”，“办公室”，“传真”，“寻呼机”......），值是任何文本格式的实际数字。

我想查询一个有两个电话号码的人：

Select * From person where
   in his phone_numbers exists phonetype = 'home' and number = '0-123-456'
   and also in his phone_numbers exists phonetype = 'mobile' and number = '9-876-421'
   (and possibly, dynamically others)
   and name = 'John'

我已经构建了一个有效的sql子查询：

select home.userId from
(
    (SELECT userId from person_phones
    where (phonetype = 'home' and number = '0-123-456'))
) as home,
(
    (SELECT userId from person_phones
    where (phonetype = 'mobile' and number = '9-876-421'))
) as mobile
where home.userId = mobile.userId

如上所述，这只是一个sql子查询。我正在我的项目中编写JPA 2.1条件查询。这看起来很奇怪。任何人都可以给我一个提示吗？

Answer 1

有一个类似的问题，使用多个内部联接而不是子查询解决它。

EntityManagerFactory emf =  Persistence.createEntityManagerFactory("Person-Test");
    EntityManager em = emf.createEntityManager();

    Map<String, String> phoneNumbers = new HashMap<>();
    phoneNumbers.put("home","0-123-456");
    phoneNumbers.put("mobile","9-876-421");

    CriteriaBuilder cb = em.getCriteriaBuilder();
    CriteriaQuery<Person> query = cb.createQuery(Person.class);
    Root<Person> personRoot = query.from(Person.class);
    query.select(personRoot);

    phoneNumbers.forEach((k, v) -> {
        MapJoin<Person, String, String> phoneNrJoinJoin = personRoot.joinMap("phoneNumbers");
        phoneNrJoinJoin.on(cb.equal(phoneNrJoinJoin.key(), k), cb.equal(phoneNrJoinJoin.value(), v));
    });

    query.where(cb.equal(personRoot.get("name"), "John"));

    List<Person> people = em.createQuery(query).getResultList();

这导致以下hibernate查询（为清晰起见，重命名为别名）

SELECT person.person_id, person.name
FROM person 
INNER JOIN person_phones a
ON person.person_id = a.userid
AND (a.phonetype = ? AND a.NUMBER = ?) 
INNER JOIN person_phones b
on person.person_id=b.userId 
and (b.phonetype=? and b.number = ? )
WHERE
person.name = ?;

返回所有提到的电话号码匹配的所有类型的元组。