CODEWITHSUNDEEP
hibernatejpaplayframework
Abhishek
Abhishek

Reputation: 6900

property mapping has wrong number of columns exception - Play-framework

I am a beginner with play framework.Again a question on JPA and mappings in play framework,

I have a student table and a mentor table bound by a one to one relationship.

Student table :

id, name, class, grade

Mentor table:

id, name, department, student_id

In the above, a mentor may or may not have a student bound to him/her. I am making the mentor Model with a one to one mapping,

@OneToOne
@JoinColumn(name="fk_student_id", referencedColumnName="id")
private student Student;

When I try to run this, I get an

A JPA error occurred (Unable to build EntityManagerFactory): property mapping has wrong number of columns: models.Mentor.student type: models.Student.

I am sure I have mapped all the Student fields as below,

Student.java

@Id @GeneratedValue(strategy=GenerationType.AUTO)
private long id;

@Column(name="name")
private String name;

@Column(name="class")
private String cls;

@Column(name="grade")
private String grade;

What am I missing here?

Thanks for your time.

Regards, Abi

Upvotes: 5

Views: 12905

Answers (2)

mosid
mosid

Reputation: 1114

Also you can use GenericModel if you want some control on your id and sequence strategy

From the play official documentation:

Custom id mapping with GenericModel:
Nothing forces you to base your entities on play.db.jpa.Model. Your JPA entities can also extend the play.db.jpa.GenericModel class. This is required if you do not want to use a Long id as the primary key for your entity.

For example, here is a mapping for a very simple User entity. The id is a UUID, the name and mail properties are required, and we use Play Validation to enforce simple business rules.

@Entity
public class User extends GenericModel {
@Id
@GeneratedValue(generator = "system-uuid")
@GenericGenerator(name = "system-uuid", strategy = "uuid")
public String id;

@Required
public String name;

@Required
@MaxSize(value=255, message = "email.maxsize")
@play.data.validation.Email
public String mail;
}

Upvotes: 0

Pere Villega
Pere Villega

Reputation: 16439

Are you sure this is working code for Play Framework? There are some differences between Play and standard JPA when creating your model. This fragment:

@OneToOne
@JoinColumn(name="fk_student_id", referencedColumnName="id")
private student Student;

is wrong. Should be something like

@OneToOne
@JoinColumn(name="fk_student_id") //removed the id reference, let JPA manage it
public Student student; //note order of class and var name

Also, you are defining an 'id' field, which is not needed when you extend from Model. Are you extending from Model?

Upvotes: 2

Related Questions