Question

我使用Criteria API创建动态查询。我有实体：

class ProcessingActivityCategory{
(...)
@OneToMany
    @JoinColumn(name = "id_kategoria_czynnosci", referencedColumnName = "id")
    private Set<ActivityCategory> activityCategories;

}

和几个嵌套实体。我只想获取这些满足以下要求的ProcessingActivityCategory.activityCategories：

    ProcessingActivityCategory.activityCategories.processActivity.personalData.company.id=x 
and ProcessingActivityCategory.activityCategories.processActivity.personalData.administrator.id=y

下面是我的实体。如何使用@Filter和@FilterDef实现此目的？我应该在哪个地方/级别使用它？对于财产 ProcessingActivityCategory.activityCategories？但是，这个地方变得更深，创造了一个条件？

ActivityCategory：

    @Entity
    @Table(schema = "odo", name = "odo_czynnosc_kategoria")
    public class ActivityCategory extends BaseEntity {

        @OneToOne
        @JoinColumn(name = "id_czynnosc", referencedColumnName = "id")
        private ProcessingActivity processActivity;
(...)

ProcessingActivity：

Data
@Entity
@Table(schema = "odo", name = "odo_czynnosc_przetwarzania")
public class ProcessingActivity extends BaseEntity {

    @OneToOne
    @JoinColumn(name = "id_zbior", referencedColumnName = "id")
    private PersonalData personalData;
(...)

个人数据：

Entity
@Table(schema = "odo", name = "odo_zbior")
@Data
public class PersonalData extends NameDescriptionEntity {

    @OneToOne
    @JoinColumn(name = "id_zbior_spolka", referencedColumnName = "id")
    private Company company;

    @OneToOne
    @JoinColumn(name = "id_zbior_rodzaj", referencedColumnName = "id")
    private DataKind dataKind;

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

    @OneToOne
    @JoinColumn(name = "id_zbior_administrator", referencedColumnName = "id")
    private Company administrator;

公司：

@Entity
@Data
@Table(schema = "odo", name = "d_odo_spolka")
public class Company{
private Long id
(...)

使用JPA的Hibernate筛选器-嵌套实体

0 个答案: