Hibernate 如何为HQL编写查询？

我有一个项目：监控学生的成绩。 我有一些实体。有：

@Entity
@Table(name = "clazzes")
public class Clazz extends BaseEntity{

    @NotEmpty
    @Column(name = "number")
    private String number;

    @NotEmpty
    @Column(name = "letter")
    private String letter;

    @JsonIgnore
    @OneToMany(cascade = CascadeType.ALL, mappedBy = "clazz", fetch = FetchType.EAGER)
    private Set<Student> students;

    @ManyToOne(fetch=FetchType.EAGER)
    @JoinColumn(name="school_id")
    private School school;

    // getters and setters
}

@Entity
@Table(name = "students")
public class Student extends Person {

    @JsonIgnore
    @ManyToMany(fetch = FetchType.EAGER)
    @JoinTable(name = "parents_students", joinColumns = @JoinColumn(name = "student_id"),
            inverseJoinColumns = @JoinColumn(name = "parent_id"))
    private List<Parent> parents;

    @NotNull
    @JsonIgnore
    @ManyToOne
    @JoinColumn(name = "clazz_id")
    private Clazz clazz;

    @OneToMany(cascade = CascadeType.ALL, mappedBy = "student", fetch = FetchType.EAGER)
    private Set<Grade> grades;

    // getters and setters
}


@Entity
@Table(name = "grades")
public class Grade extends BaseEntity{

    @ManyToOne(fetch=FetchType.EAGER)
    @JoinColumn(name = "shedule_id")
    private Shedule shedule;

    @NotNull
    @JsonIgnore
    @ManyToOne(fetch=FetchType.EAGER)
    @JoinColumn(name="student_id")
    private Student student;

    @Column(name = "task")
    private String task;

    @Column(name = "mark")
    private Integer mark;

    // getters and setters
}


@Entity
@Table(name = "shedule")
public class Shedule extends BaseEntity implements Comparable<Shedule>{

    @NotNull
    @Column(name = "date")
    @JsonFormat(shape = JsonFormat.Shape.STRING, pattern = "dd.MM.yyyy")
    @DateTimeFormat(pattern = "dd.MM.yyyy")
    private Date date;

    @NotNull
    @ManyToOne(fetch=FetchType.EAGER)
    @JoinColumn(name="period_id")
    private Period period;

    @NotNull
    @ManyToOne(fetch=FetchType.EAGER)
    @JoinColumn(name="subject_id")
    private Subject subject;

    @NotNull
    @ManyToOne(fetch=FetchType.EAGER)
    @JoinColumn(name="clazz_id")
    private Clazz clazz;

    @NotNull
    @ManyToOne(fetch=FetchType.EAGER)
    @JoinColumn(name="teacher_id")
    private Teacher teacher;

    @Column(name = "job")
    private String job;

    @JsonIgnore
    @OneToMany(cascade = CascadeType.ALL, mappedBy = "shedule", fetch = FetchType.EAGER)
    private Set<Grade> grades;

    // getters and setters
}

但是由于附加条件“with（grade.shedule.id=shedule.id）”，我得到了以下错误：

如果将使用此项目架构，是否可以编写HQL查询？还是有必要改变项目的架构？还是有必要编写本机查询

SELECT DISTINCT shedule, students, grade
      FROM Shedule shedule
      INNER JOIN shedule.clazz clazz 
      INNER JOIN clazz.students students
      LEFT JOIN students.grades grade
      WHERE grade IS NULL OR grade.shedule.id = shedule.id

java.lang.IllegalArgumentException: org.hibernate.hql.internal.ast.QuerySyntaxException: with-clause referenced two different from-clause elements [SELECT DISTINCT shedule, students, grade FROM com.vizaco.onlinecontrol.model.Shedule shedule INNER JOIN shedule.clazz clazz INNER JOIN clazz.students students LEFT JOIN students.grades grade with (grade.shedule.id = shedule.id)

SELECT DISTINCT shedule, students, grade
      FROM Shedule shedule
      INNER JOIN shedule.clazz clazz 
      INNER JOIN clazz.students students
      LEFT JOIN students.grades grade
      WHERE grade IS NULL OR grade.shedule.id = shedule.id