简体   繁体   中英

Jpa Criteria for ManyToMany and OneToOne relationships

I have 3 POJO classes - Link, LinkDetails and Tag. The relationship between Link and LinkDetails - OneToOne, between LinkDetails and Tag - ManyToMany.

How to use Jpa Criteria, find a list of links with a specified tag name ?

@Entity
public class Link extends AbstractEntity {

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private Long id;

    @Column
    private String url;

    @OneToOne(cascade = CascadeType.ALL)
    @PrimaryKeyJoinColumn
    private LinkDetails linkDetails;
}

@Entity
public class LinkDetails extends AbstractEntity {

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private Long id;    

    @Column
    private String description;

    @JoinTable(name = "link_details_2_tag", joinColumns = { @JoinColumn(name = "link_details_id")}, inverseJoinColumns = { @JoinColumn(name = "tag_id") })
    @ManyToMany(targetEntity = Tag.class, fetch = FetchType.LAZY)
    private Set<Tag> tags = new TreeSet<Tag>();

}

@Entity
public class Tag extends AbstractEntity {

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private Long id;

    @Column
    private String name;
}
@Override
public List<Link> getLinksByTag(String tag){

    CriteriaBuilder cBuilder = getEntityManager().getCriteriaBuilder();
    CriteriaQuery<Link> criteria = cBuilder.createQuery( Link.class );
    Root<Link> linkRoot = criteria.from( Link.class );
    Join<Link, LinkDetails> linkDetailsJoin = linkRoot.join(Link_.linkDetails);
    Join<LinkDetails, Tag> tagJoin = linkDetailsJoin.join(LinkDetails_.tags);
    criteria.select(linkRoot);
    criteria.where(cBuilder.equal(tagJoin.get(Tag_.name), tag));
    TypedQuery<Link> query = getEntityManager().createQuery(criteria);
    return query.getResultList();
}

The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM