多对一的 JPA 标准 API 规范答案

【问题标题】：JPA Criteria API Specification for Many to One多对一的 JPA 标准 API 规范
【发布时间】：2021-05-10 12:26:30
【问题描述】：

我有两张表 Student 和 Address。一个学生可以有多个地址。实体看起来像这样

@Entity @Table(name = "STUDENT") 
public class Student {
  @Column(name = "STUDENT_ID") Integer studentId;
  @Column(name = "FIRST_NAME") String fName;
  @Column(name = "LAST_NAME") String LName;
}

@Entity @Table(name = "ADDRESS")
public class Address {
  @Column(name = "ADDRESS_ID") Integer addressId;
  @Column(name = "STREET_NAME") String street_name;

  @ManyToOne(fetch = FetchType.LAZY, cascade = CascadeType.ALL)
  @JoinColumn(name = "STUDENT_ID")
  private Student student;
}

我需要使用 JPA Criteria 来获取基于学生 fName 和 LName 的所有地址。可以是INNER JOIN，也可以是Address表的子查询。

CriteriaBuilder builder = em.getCriteriaBuilder();
CriteriaQuery<Address> criteriaQuery = builder.createQuery(Address.class);

Root<Address> fromAddress = criteriaQuery.from(Address.class);
//Can be a JOIN or Sub-Query.
em.createQuery(criteriaQuery).getResultList();

要求是现在才加载地址表数据。

【问题讨论】：

标签： java jpa spring-data-jpa many-to-one criteriaquery

【解决方案1】：

    CriteriaBuilder builder = em.getCriteriaBuilder();
    CriteriaQuery<Address> criteriaQuery = builder.createQuery(Address.class);

    Root<Address> fromAddress = criteriaQuery.from(Address.class);
    //join
    Join<Address, Student> studentJoin = fromAddress.join("student");
    //where
    criteriaQuery.where(builder.and(
            builder.equal(studentJoin.get("fName"), fName),
            builder.equal(studentJoin.get("lName"), lName)
    ));
    //projection
    criteriaQuery.select(fromAddress);
    return em.createQuery(criteriaQuery).getResultList();

【讨论】：