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Student 类与 Resume 类型的 Object 具有 OneToOne 关联,Resume 类与 Master 类型的集合具有 OneToMany 关联,而 master 具有属性 degreeName。我想要实现的是:

选择 student.resume.masters.degreeName 在 (?,?,?,.....) 中的所有学生

它应该搜索所有集合(主)对象。

下面的代码没有编译错误,但没有给出预期的结果,请纠正我。

学生实体:

 @Entity
 public class Student {
      @OneToOne(fetch = FetchType.LAZY, mappedBy = "student", cascade = CascadeType.ALL)
      private Resume resume;
   }  

简历实体:

@Entity
public class Resume {
     @OneToMany(mappedBy="resume",cascade=CascadeType.ALL)
     private List<Master> masters=new ArrayList<>();
     @OneToOne(fetch=FetchType.LAZY)
     @PrimaryKeyJoinColumn
     private Student student;
   }  

主实体:

@Entity
public class Master {
   @Id
   @GeneratedValue(strategy=GenerationType.AUTO)
   private int id;
   @ManyToOne(fetch=FetchType.LAZY)
   @JoinColumn(name="resume_id")
   private Resume resume;
   private String degreeName;
}

道:

    Criteria studentCriteria = session.createCriteria(Student.class);
    Criteria resumeCriteria = studentCriteria.createCriteria("resume");
    Criteria mastersCriteria = resumeCriteria.createCriteria("masters");
    List<String> degreeslist = new ArrayList<>(Arrays.asList(degrees));
    //degreeList is collection of values on which the student will be searched

    if (degreeslist.size() == 1) {
        mastersCriteria.add(Restrictions.eq("degreeName",
                degreeslist.get(0)));
    } else {

        mastersCriteria.add(Restrictions.in("degreeName", degreeslist));
      }
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1 回答 1

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    Criteria mastersCriteria = getSession().createCriteria(Student.class, "student");
    mastersCriteria.setResultTransformer(Criteria.DISTINCT_ROOT_ENTITY);    
    mastersCriteria.createAlias("student.resume", "resume"); 
    mastersCriteria.createAlias("resume.masters","masters");

    if (degreeslist.size() == 1) {
          mastersCriteria.add(Restrictions.eq("masters.degreeName", degreeslist.get(0)));  
    } else {
          mastersCriteria.add(Restrictions.in("masters.degreeName", degreeslist));
    }


    List<Student> students = mastersCriteria.list();
于 2013-08-27T10:11:18.017 回答