java - 如何使用 JPA 和 Hibernate 返回一个 Map，其中键是实体属性，值是实体

Question

假设我有一个 sql 表

create table CUSTOMER
(
  first_name      VARCHAR2(30) not null,
  last_name       VARCHAR2(30) not null,
  age             VARCHAR2(30) not null,
  credit_card_num VARCHAR2(30) not null UNIQUE
);

我想要一个 Java Map，其中映射键的值为 CUSTOMER.credit_card_num map 值都是 CUSTOMER 的属性 例如:

1234134cardnumber -> Customer(first_name: Jack, age: 23, credit_card_num: 1234134cardnumber),
123faef -> Customer(first_name: Mike, age: 43, credit_card_num: 123faef)

等等...

我会接受在 JPA 或 Native Query 中执行此操作的任何方式

更新

如 https://stackoverflow.com/users/1025118/vlad-mihalcea 所提及可以通过https://vladmihalcea.com/why-you-should-use-the-hibernate-resulttransformer-to-customize-result-set-mappings/来完成

Answer 1

这个问题有两种解决方案。

假设您使用以下 Post 实体:

@Entity(name = "Post")
@Table(name = "post")
public class Post {

    @Id
    private Long id;

    private String title;

    @Column(name = "created_on")
    private LocalDate createdOn;

    public Long getId() {
        return id;
    }

    public Post setId(Long id) {
        this.id = id;
        return this;
    }

    public String getTitle() {
        return title;
    }

    public Post setTitle(String title) {
        this.title = title;
        return this;
    }

    public LocalDate getCreatedOn() {
        return createdOn;
    }

    public Post setCreatedOn(LocalDate createdOn) {
        this.createdOn = createdOn;
        return this;
    }
}

并且您的数据库中保留了以下实体:

entityManager.persist(
    new Post()
        .setId(1L)
        .setTitle("High-Performance Java Persistence eBook has been released!")
        .setCreatedOn(LocalDate.of(2016, 8, 30))
);

entityManager.persist(
    new Post()
        .setId(2L)
        .setTitle("High-Performance Java Persistence paperback has been released!")
        .setCreatedOn(LocalDate.of(2016, 10, 12))
);

entityManager.persist(
    new Post()
        .setId(3L)
        .setTitle("High-Performance Java Persistence Mach 1 video course has been released!")
        .setCreatedOn(LocalDate.of(2018, 1, 30))
);

entityManager.persist(
    new Post()
        .setId(4L)
        .setTitle("High-Performance Java Persistence Mach 2 video course has been released!")
        .setCreatedOn(LocalDate.of(2018, 5, 8))
);

entityManager.persist(
    new Post()
        .setId(5L)
        .setTitle("Hypersistence Optimizer has been released!")
        .setCreatedOn(LocalDate.of(2019, 3, 19))
);

使用 Java 8 流

以下是您如何根据问题的要求返回 map :

Map<Long, Post> postByIdMap = entityManager
.createQuery(
    "select p " +
    "from Post p ", Post.class)
.getResultStream()
.collect(
    Collectors.toMap(
        Post::getId,
        Function.identity()
    )
);

assertEquals(
    "High-Performance Java Persistence eBook has been released!",
    postByIdMap.get(1L).getTitle()
);

assertEquals(
    "Hypersistence Optimizer has been released!",
    postByIdMap.get(5L).getTitle()
);

使用 Hibernate ResultTransformer

使用 ResultTransformer 可以实现相同的目标:

Map<Long, Post> postByIdMap = (Map<Long, Post>) entityManager
.createQuery(
    "select p " +
    "from Post p ")
.unwrap(org.hibernate.query.Query.class)
.setResultTransformer(
    new ResultTransformer() {

        Map<Long, Post> result = new HashMap<>();

        @Override
        public Object transformTuple(Object[] tuple, String[] aliases) {
            Post post = (Post) tuple[0];
            result.put(
                post.getId(),
                post
            );
            return tuple;
        }

        @Override
        public List transformList(List collection) {
            return Collections.singletonList(result);
        }
    }
)
.getSingleResult();

assertEquals(
    "High-Performance Java Persistence eBook has been released!",
    postByIdMap.get(1L).getTitle()
);

assertEquals(
    "Hypersistence Optimizer has been released!",
    postByIdMap.get(5L).getTitle()
);