Foreign key mapping inside Embeddable class
Asked Answered
D

1

14

I am using eclipselink for JPA. I have an entity, which has a composite key fabricated out of two fields. Following is my Embeddable primary key class' fields(members).

    @Embeddable
    public class LeavePK {
       @ManyToOne(optional = false)
       @JoinColumn(name = "staffId", nullable = false)
       private Staff staff;
       @Temporal(TemporalType.TIMESTAMP)
       private Calendar date;
       //setters and getters
    }

My entity is going to hold leave data related to a staff, so I am trying to combine staff object and leave date to produce composite key. Apart from my logic, it is not allowing me to have a foreign key mapping inside embeddable class. When I try to use JPA tools--> Generate Tables From Entity, it gives error as below, which explains, but I am not getting it.

org.eclipse.persistence.exceptions.ValidationException
Exception Description: The mapping [staff] from the embedded ID class [class rs.stapp.entity.LeavePK] is an invalid mapping for this class. An embeddable class that is used with an embedded ID specification (attribute [leavePK] from the source [class rs.stapp.entity.Leave]) can only contain basic mappings. Either remove the non basic mapping or change the embedded ID specification on the source to be embedded.

Does it mean, I cannot have a key(from composite key) which is also a foreign key. Is there a alternative way to accomplish this ERM? Please help. Thanks

Drawtube answered 9/4, 2012 at 18:57 Comment(0)
E
18

Don't put relationships into ID classes, neither for @IdClass or @EmbeddedId ones. An @Embeddable class may only include the annotations @Basic, @Column, @Temporal, @Enumerated, @Lob, or @Embedded. Everything else is provider-specific syntax (e.g. Hibernate allows this, but since you're using EclipseLink, which is the JPA RI, I doubt this is what you want).

Here's an example JPA PK/FK mapping:

@Entity
@Table(name = "Zips")
public class Zip implements Serializable
{
    @EmbeddedId
    private ZipId embeddedId;

    @ManyToOne
    @JoinColumn(name = "country_code", referencedColumnName = "iso_code")
    private Country country = null;

    ...
}

@Embeddable
public class ZipId implements Serializable
{
    @Column(name = "country_code")
    private String countryCode;

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

    ...
}

HTH

Eccentricity answered 10/4, 2012 at 4:31 Comment(1)
This is only correct for JPA 1.0, since then JPA 2.0 has allowed references in Embeddables.Botany

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