java加入条件查询出现错误“无法定位属性”

Question

我有两张桌子和两门课。我必须连接两个表并获得特定字段

class Student extends Parent{
  
  Long id;
  @Cache(usage = CacheConcurrencyStrategy.READ_WRITE)
  @OneToMany(fetch = FetchType.LAZY, mappedBy = courses.student, cascade = CascadeType.ALL)
    Set<Courses> coursesList = new HashSet<>();
 }
 
 @Table(name = "courses", indexes = {
    @Index(name = "id", columnList = Courses.id, unique = true),
    @Index(name = "student_course_fk", columnList = "student_fk"),
 class Courses extends Parent{
     Long id;
     @ManyToOne
     @JoinColumn(name = "student_fk")
     Student student;
   }
 

如果我质疑：-

CriteriaBuilder cb=entityManager.getCriteriaBuilder();
CriteriaQuery<Student> q = cb.createQuery(Student.class);
Root<Student> c = q.from(Student.class);
Join<Student,Courses> lineJoin  = root.join("Courses" 
,JoinType.INNER);
lineJoin.on(criteriaBuilder.equal(root.get(Courses.student),
root.get(student.id)));

我得到“无法在此ManagedType[Parent]上找到名为[student]的属性” 有人能帮我加入吗。我知道我把这两张桌子放在一起是做错了什么

Answer 1

我将改变以下几点：

@OneToMany(fetch = FetchType.LAZY, mappedBy = courses.student,...

@OneToMany(fetch = FetchType.LAZY, mappedBy = student,...

必须指明在参数类型（Set）中定义的目标类的参数

Join<Student,Courses> lineJoin  = root.join("Courses",JoinType.INNER);

Join<Student,Courses> lineJoin  = root.join("coursesList",JoinType.INNER);

如上所述，join参数必须与参数的名称匹配

lineJoin.on(criteriaBuilder.equal(root.get(Courses.student)

lineJoin.on(criteriaBuilder.equal(root.get("student")

或者

lineJoin.on(criteriaBuilder.equal(root.get(Courses_.student)

要使用get，必须传递一个带有字段名称的字符串作为参数，或者使用元模型（以下划线结尾）