如何将MySQL查询转换为HQL查询？

Question

I am new to HQL. This is a mysql Query. I need to convert it into HQL query. How to do that any suggestions please?

    `SELECT STUDENT.ID, STUDENT.NAME, STUDENT.GRADE_ID, STUDENT.CLASS, GRADE.NAME FROM
     STUDENT INNER JOIN GRADE ON STUDENT.GRADE_ID = GRADE.GRADE_ID`

STUDENT[ID, NAME, GRADE_ID, CLASS]

GRADE[GRADE_ID, GRADE_NAME]

The resultset of the above query will be something like ID, NAME, GRADE_NAME, CLASS.

---

Final Update:

I have 2 tables STUDENT[ID, NAME, GRADE_ID, CLASS] GRADE[GRADE_ID, GRADE_NAME]

With the SQL query SELECT STUDENT.ID, STUDENT.NAME, STUDENT.GRADE_ID, STUDENT.CLASS, GRADE.NAME FROM STUDENT INNER JOIN GRADE ON STUDENT.GRADE_ID = GRADE.GRADE_ID I used to show a New table STUDENT[ID, NAME, GRADE_NAME, CLASS] was working in SQL. For this function there is no input, return type is a list(all the records of the result table)

This is what i want to do in HQL or JPQL , How to get the List of Objects where, Properties of the Object were id, name, gradename, class.

How to do it with HQL.

Answer 1

You wouldn't do that in Hibernate. The whole point of using hibernate is that you are dealing with objects.

So I guess your Student class would have a List of type Grade

@Entity
public class Student{
    // accessors and id omitted
    @OneToMany(mappedBy="student")
    private List<Grade> grades;
}

@Entity
public class Grade{
    // accessors and id omitted
    @ManyToOne
    private Student student;
}

You would look up the grade via session.get() and then do grade.getStudent() to access the student:

Grade grade = (Grade) session.get(Grade.class, gradeId);
Student student = grade.getStudent();

HQL queries are designed for Scenarios that are too complex for these lookup methods.

Edit: I just realized that the question is tagged jpa . Then of course you would not be using HQL, but JPQL instead. And also you'd be using this code:

Grade grade = entityManager.find(Grade.class, gradeId); // no cast needed with JPA 2
Student student = grade.getStudent();

---

Edit: given the requirements you added in your comments I'd change the data model to something like this (ids omitted):

@Entity
public class Student{
    public List<Course> getCourses(){
        return courses;
    }
    public void setCourses(List<Course> courses){
        this.courses = courses;
    }
    @OneToMany(mappedBy="student")
    private List<Course> courses;
}

@Entity
public class Course{
    @ManyToOne(optional=false)
    private Student student;
    @ManyToOne(optional=false)
    private Grade grade;
    public void setStudent(Student student){
        this.student = student;
    }
    public Student getStudent(){
        return student;
    }
    public Grade getGrade(){
        return grade;
    }
    public void setGrade(Grade grade){
        this.grade = grade;
    }
}

@Entity
public class Grade{
    @OneToMany(mappedBy="grade")
    private Set<Course> courses;
    public void setCourses(Set<Course> courses){
        this.courses = courses;
    }
    public Set<Course> getCourses(){
        return courses;
    }
}

And I'd query like this:

Grade grade = entityManager.find(Grade.class, 1L);
List<Student> studentsWithThisGrade = new ArrayList<Student>();
for(Course course : grade.getCourses()){
    studentsWithThisGrade.add(course.getStudent());
}

(But Grade should probably not be an entity, but either a numeric value or an enum.)

---

It turns out that the OP uses plain hibernate after all, no JPA. So whenever you see entityManager.find() above, replace that in your mind with session.get() :-)

Answer 2

Try this.

public List<Student> getByGradeId(EntityManager entityManager, Grade grade) {
   Query query = entityManager.createQuery("from Student where grade=:grade");
   query.setParameter("grade", grade);
   return (List<Student>) query.getResultList();
}

You'll need a try/catch in there too.

Read section 3.4 of this !