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Hibernate/JPA: could not set a field value by reflection setter

My JPA/Hibernate odyssey continues...

I am trying to work around this issue, and so I have had to define primitive @Ids in my class that uses 3 entity fields as a composite key. This seems to get me a bit further, but now I'm getting this when persisting:

javax.persistence.PersistenceException: org.hibernate.PropertyAccessException: could not set a field value by reflection setter of com.example.model.LanguageSkill.stafferId

Here's my composite class:

public class LanguageSkill implements Serializable
{
    @Id
    @GeneratedValue (strategy = GenerationType.IDENTITY)
    @Column(name = "Staffer_ID")
    private Long stafferId;

    @Id
    @ManyToOne(cascade = CascadeType.ALL)
    @MapsId(value = "stafferId")
    private Staffer staffer;

    @Id
    @GeneratedValue (strategy = GenerationType.IDENTITY)
    @Column(name = "Language_ID")
    private Long languageId;

    @ManyToOne
    @MapsId(value= "languageId")
    private Language language;

    @Id
    @GeneratedValue (strategy = GenerationType.IDENTITY)
    @Column(name = "Language_Proficiency_ID")
    private Long languageProficiencyId;

    @ManyToOne
    @MapsId(value= "languageProficiencyId")
    private LanguageProficiency languageProficiency;
}

I do have proper getters and setters (IDE-generated) for both the primitives as well as the entities.

Here are my libs. I'm not totally convinced that I'm using a compatible set of persistence libraries (references to a cookbook detailing how to properly mix-and-match these would be highly appreciated.)

  • Hibernate 3.5.6-SNAPSHOT
  • hibernate-jpamodelgen 1.1.0.CR1
  • hibernate-validator 3.1.0.GA
  • MySQL 5.1.6
  • jsr250-api 1.0
  • javax.validation validation-api 1.0.0.GA

Wow, it's frustrating. 3 days now full time trying to solve various issues like this just for basic ORM. I feel defective. :-(

like image 554
George Armhold Avatar asked Nov 25 '22 15:11

George Armhold


1 Answers

It seems a correct code. I had problem with this exception when I used Blob[]

@Lob
@Column(name="DOCUMENTO",nullable=false)
private Blob[] documento;

But changing by Byte[], I solved this problem.

I have only a occurrence, looking Oracle data types, I have seen this LONG is Character data of variable length (A bigger version the VARCHAR2 datatype).

I assume that your ID is a Integer....Why not change Long by Integer? You must remember that it only accepts primitive types.

This is my code and it works fine:

@Id
@SequenceGenerator(sequenceName="SQ_DOCUMENTO",name="seqDocumento")
@GeneratedValue(strategy=GenerationType.SEQUENCE,generator="seqDocumento")
private Integer idDocumento;

I use Hibernate 3.5.6-final, Spring 3.0.4, Junit 4 and Oracle 11g.

like image 155
Rafael Ruiz Tabares Avatar answered Feb 02 '23 00:02

Rafael Ruiz Tabares