我需要搜索包含字符串参数的数据
我有两个实体:
@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标准的搜索API,它可以很好地处理不同的字段和join(ManyToOne),但是在这个例子中,我需要找到Referentiel,它包含referentielDigital中的字符串。
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“在”,我想有“喜欢”,以获得良好的结果。
对于Extension对象,我没有“like”方法。
请问我怎样才能用这个加入来做喜欢的请求呢?
1条答案
按热度按时间bz4sfanl1#
我找到解决方案与JB Nizet帮助: