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PO
 

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    <p>You wouldn't do that in Hibernate. The whole point of using hibernate is that you are dealing with objects.</p> <p>So I guess your Student class would have a List of type Grade</p> <pre><code>@Entity public class Student{ // accessors and id omitted @OneToMany(mappedBy="student") private List&lt;Grade&gt; grades; } @Entity public class Grade{ // accessors and id omitted @ManyToOne private Student student; } </code></pre> <p>You would look up the grade via <a href="http://docs.jboss.org/hibernate/core/3.6/javadocs/org/hibernate/Session.html#get%28java.lang.Class,%20java.io.Serializable%29" rel="nofollow noreferrer">session.get()</a> and then do <code>grade.getStudent()</code> to access the student:</p> <pre><code>Grade grade = (Grade) session.get(Grade.class, gradeId); Student student = grade.getStudent(); </code></pre> <p>HQL queries are designed for Scenarios that are too complex for these lookup methods.</p> <p>Edit: I just realized that the question is tagged <code>jpa</code>. Then of course you would not be using HQL, but JPQL instead. And also you'd be using this code:</p> <pre><code>Grade grade = entityManager.find(Grade.class, gradeId); // no cast needed with JPA 2 Student student = grade.getStudent(); </code></pre> <hr> <p>Edit: given the requirements you added in your comments I'd change the data model to something like this (ids omitted):</p> <pre><code>@Entity public class Student{ public List&lt;Course&gt; getCourses(){ return courses; } public void setCourses(List&lt;Course&gt; courses){ this.courses = courses; } @OneToMany(mappedBy="student") private List&lt;Course&gt; 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&lt;Course&gt; courses; public void setCourses(Set&lt;Course&gt; courses){ this.courses = courses; } public Set&lt;Course&gt; getCourses(){ return courses; } } </code></pre> <p>And I'd query like this:</p> <pre><code>Grade grade = entityManager.find(Grade.class, 1L); List&lt;Student&gt; studentsWithThisGrade = new ArrayList&lt;Student&gt;(); for(Course course : grade.getCourses()){ studentsWithThisGrade.add(course.getStudent()); } </code></pre> <p>(But Grade should probably not be an entity, but either a numeric value or an enum.)</p> <hr> <p>It turns out that the OP uses plain hibernate after all, no JPA. So whenever you see <code>entityManager.find()</code> above, replace that in your mind with <code>session.get()</code> :-)</p>
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    1. POHow to convert MySQL query into HQL query?
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    1. USSean Patrick Floyd
    1. USSean Patrick Floyd
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    1. POHow to convert MySQL query into HQL query?
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    1. COI rather thought that many Students could shared a Grade. So he could be trying to find all students that got a "B" for example. IN which case, the query is appropriate.
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      1. USTony Ennis
    2. CO@SEANIZER I want to get the whole table using a join between the two tables, where GRADE_ID replaced by GRADE_NAME. the resultset wont be having the GRADE_ID column.
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      1. USIBM'er
    3. COThe problem is that you are still thinking in tables, whereas you should be thinking in objects if you use hibernate and / or jpa. Read this: http://en.wikipedia.org/wiki/Object-relational_impedance_mismatch
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      1. USSean Patrick Floyd
 

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