Scala with JPA

In my quest to make Scala blend with JSF and JPA, I’ve been trying to get an app working where I interact with the database while using Scala entities. Obviously, as a first step, I got a simple JPA project working, where my User class looked like:

import javax.persistence.*;

@Table(name = "user")
@NamedQueries({@NamedQuery(name = "User.findAll", query = "SELECT u FROM User u"),
@NamedQuery(name = "User.findById", query = "SELECT u FROM User u WHERE = :id"), 
@NamedQuery(name = "User.findByFirstname", query = "SELECT u FROM User u WHERE u.firstname = :firstname"),
@NamedQuery(name = "User.findByLastname", query = "SELECT u FROM User u WHERE u.lastname = :lastname")})
public class User implements Serializable
    private static final long serialVersionUID = 1L;
    @Basic(optional = false)
    @Column(name = "ID")
    private Integer id;
    @Column(name = "FIRSTNAME")
    private String firstname;
    @Column(name = "LASTNAME")
    private String lastname;

    public User() {

    public User(Integer id) { = id;

    public User(Integer id, String firstname, String lastname) { = id;
        this.firstname= firstname;
        this.lastname = lastname;

    public Integer getId() {
        return id;

    public void setId(Integer id) { = id;

    public String getFirstname() {
        return firstname;

    public void setFirstname(String firstname) {
        this.firstname = firstname;

    public String getLastname() {
        return lastname;

    public void setLastname(String lastname) {
        this.lastname = lastname;

So once I got this to work, my next step was to write the same thing in Scala, I wrote it like:

import scala.reflect._
import javax.persistence._

class User(@Id @Column{val name="ID"} @BeanProperty val id:Integer,
           @Column{val name="firstname"} @BeanProperty val firstname: String,
           @Column{val name="lastname"} @BeanProperty val lastname:String)


To make this work from my Java code, I made sure I pointed to my scala User class instead of Java, this includes the persistence.xml file as well. When I tried to run this, I got the following error:

Exception [EclipseLink-0] (Eclipse Persistence Services – 1.0.1 (Build 20080905) ): org.eclipse.persistence.exceptions.IntegrityException Descriptor Exceptions:

Exception [EclipseLink-63] (Eclipse Persistence Services – 1.0.1 (Build 20080905 )): org.eclipse.persistence.exceptions.DescriptorException Exception Description:

The instance creation method [scalajpa.User.], with no parameters, does not exist, or is not accessible.

Internal Exception: java.lang.NoSuchMethodException: scalajpa.User.() Descriptor: RelationalDescriptor(scalajpa.User –> [DatabaseTable(USER)])

So I realised I needed a default constructor for my User method, so I changed my scala class to:

import scala.reflect._
import javax.persistence._

class User
    @Column{val name="ID"}
    @BeanProperty var id:Integer= _

    @Column{val name="firstname"}
    @BeanProperty var firstname: String = _

    @Column{val name="lastname"}
    var lastname:String = _

This worked out perfect. Talk about missing the obvious :-). I realised I had to make this change when I looked here.

While I really like the compactness I got out of it, I’m not too thrilled about having to create a new Scala project, where I need all the EclipseLink libraries on the classpath, then build the Scala project into a jar and put it on the classpath of the JPA application. The whole process would be much easier if I could just create a Scala class right inside my JPA application and have it accessible from other java classes. I tried this and unfortunately the java classes did not recognize the Scala class :(.

As a next step, I plan to get a JSF + JPA application working with whatever help they can get from Scala.


9 thoughts on “Scala with JPA

  1. I would like to map an Enum with eclipselink. I saw examples how I’m able to do this with hibernate. Can anybody give me a hint ?


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