JPA 存储库属性表达式

Question

在我的 Spring Boot 应用程序（基于 JHipster 4）中，我正在尝试使用属性表达式通过相关实体的属性过滤我的查询，如 Spring 文档中所述： https://docs.spring.io/spring-data/jpa/docs/current/reference/html/#repositories.query-methods.query-property-expressions

我想获取所有约会，在他们的 AppointmentSchedules 中有一个特定的 EmployeeAccount，但只收到一个空列表。

@Repository
public interface AppointmentRepository extends JpaRepository<Appointment, Long> {

        public List<Appointment> findByScheduledAppointmentsEmployeeAccountsId(Long id);
}

相关实体及其关系：

@Entity
@Table(name = "appointment")
@Cache(usage = CacheConcurrencyStrategy.NONSTRICT_READ_WRITE)
public class Appointment implements Serializable {

    @OneToMany(mappedBy = "appointments")
    @JsonIgnore
    @Cache(usage = CacheConcurrencyStrategy.NONSTRICT_READ_WRITE)
    private Set<AppointmentSchedule> scheduledAppointments = new HashSet<>();

...

}

@Entity
@Table(name = "appointment_schedule")
@Cache(usage = CacheConcurrencyStrategy.NONSTRICT_READ_WRITE)
public class AppointmentSchedule implements Serializable {

    @ManyToMany
    @Cache(usage = CacheConcurrencyStrategy.NONSTRICT_READ_WRITE)
    @JoinTable(name = "appointment_schedule_employee_accounts", joinColumns = @JoinColumn(name = "appointment_schedules_id", referencedColumnName = "id"), inverseJoinColumns = @JoinColumn(name = "employee_accounts_id", referencedColumnName = "id"))
    private Set<EmployeeAccount> employeeAccounts = new HashSet<>();

...

}

@Entity
@Table(name = "employee_account")
@Cache(usage = CacheConcurrencyStrategy.NONSTRICT_READ_WRITE)
public class EmployeeAccount implements Serializable {

    @Id
    @GeneratedValue(strategy = GenerationType.SEQUENCE, generator = "sequenceGenerator")
    @SequenceGenerator(name = "sequenceGenerator")
    private Long id;

...

}

执行查询时的休眠控制台输出：

Hibernate: select appointmen0_.id as id1_1_, appointmen0_.institution_id as institut2_1_, appointmen0_.user_account_id as user_acc3_1_ from appointment appointmen0_ where appointmen0_.user_account_id=?

如果有人能告诉我我在这里做错了什么，我将不胜感激。

编辑： 请注意，Appointment 表不包含 AppointmentSchedules 列，因为它是 OneToMany 关系。也许这就是原因？我以为 JPA 会为我处理这个逻辑......

表结构：

APPOINTMENT;
ID      ...

APPOINTMENT_SCHEDULE;
ID      APPOINTMENTS_ID    ...

APPOINTMENT_SCHEDULE_EMPLOYEE_ACCOUNTS;
EMPLOYEE_ACCOUNTS_ID    APPOINTMENT_SCHEDULES_ID

EMPLOYEE_ACCOUNT;
ID    ...

编辑 2： 我能够解决这个问题。这是一个经典的 PEBCAK 错误，因为我不小心在我的服务类 facepalm 中调用了一个类似命名的方法。非常感谢所有帮助过我的人！ Gaurav Srivastav 和 Dean 的答案都有效，但不是必需的，因为问题中的原始代码已经有效

Answer 1

在引用嵌套字段之前必须使用下划线。试试：

public List<Appointment> findByScheduledAppointments_EmployeeAccounts_Id(Long id)

Answer 2

是的，将单词 In 添加到您的方法中，并提供 Set<Long> 作为employeeAccountIds。

Set<Person> findByScheduledAppointmentsEmployeeAccountsIdIn(Set<Long> employeeIds);