与OneToOne一样的JPA标准加入

Question

我需要搜索包含字符串参数的数据

我有2个实体：

@Entity
@Table(name = "referentiel_digital")
@Cache(usage = CacheConcurrencyStrategy.NONSTRICT_READ_WRITE)
public class ReferentielDigital implements Serializable {

    private static final long serialVersionUID = 1L;

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

    @NotNull
    @Size(max = 200)
    @Column(name = "libelle_commercial", length = 200, nullable = false)
    private String libelleCommercial;

    @Size(max = 1000)
    @Column(name = "description_courte", length = 1000)
    private String descriptionCourte;

    @Size(max = 1000)
    @Column(name = "description_longue", length = 1000)
    private String descriptionLongue;

    @OneToOne
    @JoinColumn(unique = true)
    private Referentiel reference;

和

@Entity
@Table(name = "referentiel")
@Cache(usage = CacheConcurrencyStrategy.NONSTRICT_READ_WRITE)
public class Referentiel implements Serializable {

    private static final long serialVersionUID = 1L;

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

    @NotNull
    @Column(name = "uuid", nullable = false)
    private UUID uuid;

    ....

    @OneToOne(mappedBy = "reference")
    @JsonIgnoreProperties("referentiels")
    private ReferentielDigital digital;

我开发了一个使用JPA Criteria的seach API。 它适用于不同的领域，并加入（ManyToOne），但在这种情况下，我需要找到Referentiel，它在例如referentielDigital.descriptionLongue中包含字符串。

我试试这个：

        Join<Referentiel, ReferentielDigital> digital = root.join("digital");
        Expression<String> exp1 = digital.get("libelleCommercial");


        Predicate p1 = exp1.in("%" + criteria.getLibelle() + "%");

        Expression<String> exp2 = digital.get("descriptionCourte");
        Predicate p2 = exp1.in("%" + criteria.getLibelle() + "%");

        Expression<String> exp3 = digital.get("descriptionLongue");
        Predicate p3 = exp1.in("%" + criteria.getLibelle() + "%");

        predicates.add(cb.or(p1, p2, p3));

它不起作用，因为请求包含“in”，并且我想要“喜欢”以获得好的结果。

使用Extension对象，我没有“喜欢”的方法。

请问如何通过此次加入请求谎言请求？

Answer 1

我在JB Nizet帮助中找到了解决方案：

        Join<Referentiel, ReferentielDigital> digital = root.join("digital");
        Expression<String> exp1 = digital.get("libelleCommercial");

        Predicate p1 = cb.like(exp1, "%" + criteria.getLibelle() + "%");

        Expression<String> exp2 = digital.get("descriptionCourte");
        Predicate p2 = cb.like(exp2, "%" + criteria.getLibelle() + "%");

        Expression<String> exp3 = digital.get("descriptionLongue");
        Predicate p3 = cb.like(exp3, "%" + criteria.getLibelle() + "%");

        predicates.add(cb.or(p1, p2, p3));