如何在 JPA 标准中使用投影和 where 子句

Question

我有实体人

@Entity(name = "Person")
public class Person {
    @Id
    @GeneratedValue
    private Long id;
    private String name;
    @OneToMany(cascade = CascadeType.ALL, fetch = FetchType.LAZY, mappedBy = "person")
    private Set<Phone> phones=new HashSet<Phone>();
    public Person() {
    }
    public Person(String name) {
        this.name = name;
    }

广告实体电话 ：

@Entity(name = "Phone")
public class Phone {
    @Id
    @GeneratedValue
    private Long id;
    @Column(name = "`number`")
    private String number;
    @ManyToOne(fetch = FetchType.LAZY)
    @JoinColumn(name = "person_id", nullable = false)
    private Person person;
    public Phone() {
    }

它们具有一对多的关系。现在我想在jpa标准中构建这样的查询：

select p.phones from person p join phone ph where p.name = :name;

所以我想从人名是参数的 Person 实体中提取Set<Phone> phones。

我写了这个jpa标准查询：

CriteriaBuilder builder = session.getCriteriaBuilder();
        CriteriaQuery<Person> query = builder.createQuery(Person.class);
        Root<Person> root = query.from(Person.class);
        CriteriaQuery<Person> where = query.where(builder.equal(root.get("name"), "Mary Dick"));
        CompoundSelection<Set> projection = builder.construct(Set.class, root.get("phones"));
        where.select(projection); //compile error: The method select(Selection<? extends Person>) in the type CriteriaQuery<Person> is not applicable for the arguments (CompoundSelection<Set>)
    }

但它给出了编译错误：

The method select(Selection<? extends Person>) in the type CriteriaQuery<Person> is not applicable for the arguments (CompoundSelection<Set>)

怎么对？我需要元模型类吗？

Answer 1

CompoundSelection<Y> construct(Class<Y> result, Selection<?>... terms)

仅当查询涉及某些投影时，此方法才有用，这些投影未完全由单个实体类封装。如果是这种情况，第一个参数将是自定义 POJO 类(具有合适的构造函数(，其字段对应于查询的 select 子句。

在这种情况下，所选内容已经是实体类的一部分。因此，您可以简单地选择所需的字段。

 CriteriaQuery<Person> query = builder.createQuery(Person.class);
 Root<Person> root = query.from(Person.class);
 query.where(builder.equal(root.get("name"), "Mary Dick"));
 query.select(root.get("phones"));

上面的查询将返回人员列表。但是，如果您只查找可迭代的手机列表，请尝试使用稍微不同的查询。

select ph from phone ph join ph.person p where p.name = :name;

及其等效的 CriteriaQuery：

CriteriaBuilder builder = session.getCriteriaBuilder();
CriteriaQuery<Phone> query = builder.createQuery(Phone.class);
Root<Phone> root = query.from(Phone.class);
Join<Phone, Person> join = root.join(root.get("person"))            
query.where(builder.equal(join.get("name"), "Mary Dick"));