Question

鉴于以下两个实体：

public class Performer {

    @Id
    private Long id;

    private String name;

    @ManyToMany(fetch = FetchType.LAZY, cascade = CascadeType.ALL)
    @JoinTable(name = "performers_characters",
            joinColumns = @JoinColumn(name = "performer_id", nullable = false, updatable = false),
            inverseJoinColumns = @JoinColumn(name = "character_id", nullable = false, updatable = false))
    private Set<Character> characters = Sets.newHashSet();
}

和

public class Character {

    @Id
    @Column(nullable = false)
    private Long id;

    @Column(nullable = false)
    private String name;

    private Boolean alive;

    private String placeOfBirth;

    @ManyToMany(mappedBy = "characters")
    private Set<Performer> performers = Sets.newHashSet();
}

我想使用Criteria Builder创建查询，它将生成等效的SQL：

SELECT performer0_.id AS id1_11_0_,
       character2_.id AS id1_0_1_,
       performer0_.name AS name7_11_0_,
       character2_.name AS name12_0_1_,
       characters1_.character_id AS character_id2_12_0__
FROM performer performer0_
INNER JOIN performers_characters characters1_ ON performer0_.id=characters1_.performer_id
INNER JOIN character character2_ ON characters1_.character_id=character2_.id

基本上，应该选择Performer中的所有内容，但是从Character中，只应选择id和name。

目前我遇到以下情况：

CriteriaBuilder criteriaBuilder = entityManager.getCriteriaBuilder();
CriteriaQuery<Performer> baseCriteriaQuery = criteriaBuilder.createQuery(Performer.class);
Root<Performer> baseRoot = baseCriteriaQuery.from(Performer.class);
baseCriteriaQuery.select(baseRoot);
baseRoot.fetch("characters");
TypedQuery<Performer> baseTypedQuery = entityManager.createQuery(baseCriteriaQuery);
List<Performer> baseEntityList = baseTypedQuery.getResultList();

这项工作很好，除了来自Character和Performer之外，所有列都被选中。

Answer 1

使用JPA，只有在您指定不想加载为懒惰的字段时才可以这样做。

@Basic(fetch=FetchType.LAZY)
private String placeOfBirth;

@Basic(fetch=FetchType.LAZY)
private Boolean alive;

但问题是你为什么要这样做？这两个字段都是小项目，而LAZY则有更多SQL调用来获取数据。

JPA CriteriaBuilder：在连接的实体上获取特定列

1 个答案: