Spring Data JPA-规范查询

Question

有一个使用属性表达式的查询：https://docs.spring.io/spring-data/jpa/docs/current/reference/html/#repositories.query-方法。查询属性表达式：

List<Pupil> findByPupilInClassRoomTeacherName(String name);

试图通过规范执行查询：

public final class PupilSpecification {
    public static Specification<Pupil> findPupilByTeacherSpecification(String name) {
        return new Specification<Pupil>() {
            @Override
            public Predicate toPredicate(Root<Pupil> root, CriteriaQuery<?> criteriaQuery, CriteriaBuilder criteriaBuilder) {
                root = criteriaQuery.from(Pupil.class);
                Join<Pupil, PupilInClassRoom> pupil_PupilInClassRoomJoin = root.join("pupil_id");
                Join<Teacher, PupilInClassRoom> teacher_PupilInClassRoomJoin = pupil_PupilInClassRoomJoin.join("teacher_id");
                return criteriaBuilder.equal(teacher_PupilInClassRoomJoin.get("name"),name);
            }
        };
    }
}

我得到了一个错误：

org.springframework.dao.InvalidDataAccessApiUsageException: Unable to locate Attribute  with the the given name [pupil_id] on this ManagedType [com.SchoolJournal.SpringHibernate.model.BaseModel]; nested exception is java.lang.IllegalArgumentException: Unable to locate Attribute  with the the given name [pupil_id] on this ManagedType [com.SchoolJournal.SpringHibernate.model.BaseModel]

我做错了什么或忘记添加了什么？

查询：按教师姓名显示所有学生

基地实体：

@Entity
@Inheritance(strategy = InheritanceType.JOINED)
abstract class BaseModel {

@Id
@Column(name = "id")
@Getter
@GeneratedValue(strategy = GenerationType.AUTO)
private long id;

protected BaseModel() {
  }
}

实体教师：

@Entity
@Table(name = "teacher")
public final class Teacher extends BaseModel {

@OneToOne
@JoinColumn(name = "teacher_id")
private PupilInClassRoom pupilInClassRoom;

@Column(nullable = false)
@Getter
@Setter
private String name;

@Column(nullable = false)
@Getter
@Setter
private String surname;

@Column(nullable = false)
@Getter
@Setter
private String discipline;

public Teacher() {
  }
}

实体学生：

@Entity
@Table(name = "pupil")
public final class Pupil extends BaseModel {

@OneToOne
@JoinColumn(name = "pupil_id")
private PupilInClassRoom pupilInClassRoom;

@Column(nullable = false)
@Getter
@Setter
private String name;

@Getter
@Setter
@Column(nullable = false)
private String surname;

public Pupil() {

  }
}

实体教室：

@Entity
@Table(name = "class_room")
public final class ClassRoom extends BaseModel {

@OneToOne
@JoinColumn(name = "class_Room_id")
private PupilInClassRoom pupilInClassRoom;

@Column(nullable = false)
@Getter
@Setter
private String name;

public ClassRoom() {
   }
}

实体教室：

@Entity
@Table(name = "pupil_in_class_room")
public final class PupilInClassRoom  extends BaseModel {

@OneToOne(mappedBy = "pupilInClassRoom")
@Getter
@Setter
private Pupil pupil;

@OneToOne(mappedBy = "pupilInClassRoom")
@Getter
@Setter
private ClassRoom classRoom;

@OneToOne(mappedBy = "pupilInClassRoom")
@Getter
@Setter
private Teacher teacher;

public PupilInClassRoom() {

}

Answer 1

问题的解决方案：

public final class PupilSpecification {
        public static Specification<Pupil> findPupilByTeacherSpecification(String name) {
            return new Specification<Pupil>() {
                @Override
                public Predicate toPredicate(Root<Pupil> root, CriteriaQuery<?> criteriaQuery, CriteriaBuilder criteriaBuilder) {
                    root = criteriaQuery.from(Pupil.class);
                    Join<Pupil, PupilInClassRoom> pupil_PupilInClassRoomJoin = root.join("pupilInClassRoom");
                    Join<PupilInClassRoom, Teacher> teacher_PupilInClassRoomJoin = pupil_PupilInClassRoomJoin.join("teacher");
                    return criteriaBuilder.equal(teacher_PupilInClassRoomJoin.get(Teacher_.Name),name);
                }
            };
        }
    }

并添加元模型教师_

@Generated(value = "org.hibernate.jpamodelgen.JPAMetaModelEntityProcessor")
@StaticMetamodel(Teacher.class)
public abstract class Teacher_ {
    public static volatile SingularAttribute<Teacher,String> name;
    public static volatile SingularAttribute<Teacher,String> surmane;
    public static volatile SingularAttribute<Teacher,String> discipline;
    public static volatile SingularAttribute<BaseModel,Integer> id;

    public static final String Name = "name";
    public static final String Surname = "surname";
    public static final String Discipline = "discipline";
    public static final String Id = "id";
}

通话规格：

pupilRepository.findAll(PupilSpecification.findPupilByTeacherSpecification("Alla"))

Answer 2

Join<Pupil, PupilInClassRoom>   pupil_PupilInClassRoomJoin   =
        root.join("pupilInClassRoom");
Join<PupilInClassRoom, Teacher> teacher_PupilInClassRoomJoin = 
        pupil_PupilInClassRoomJoin.join("teacher");

root.join（"pupil_id"）-

<代码>加入

学生加入。加入（“教师id”）-

上面的更新是我猜测做你想做的事情需要的。可以试试吗？