在单个表继承类型中查找特定类型的元素

问题描述 投票:0回答:1

想象以下实体层次:

@Entity
@Inheritance(strategy = InheritanceType.SINGLE_TABLE)
@DiscriminatorColumn(name = "type", columnDefinition = "varchar(60)")
abstract class Resource {

}


@Entity
@DiscriminatorValue("resource1")
class Resource1 extends Resource {
  @Embedded
  private Property property1;
}

@Entity
@DiscriminatorValue("resource2")
class Resource2 extends Resource {
  @Embedded
  private Property property2;
}

@Entity
@DiscriminatorValue("resource3")
class Resource3 extends Resource {
  @Embedded
  private Property property3;
}

@Entity
@DiscriminatorValue("resource4")
class Resource4 extends Resource {
  @Embedded
  private Property property4;
}

@Entity
class EntityUsingResource {
  @OneToMany(...)
  @JoinColumn(...)
  private List<Resource> resources;
}

我正在尝试创建一个UI来搜索EntityUsingResource,并且能够过滤具有具有特定属性资源的元素。

因此,您可以在GUI的搜索字段中为property4键入一个值,它会过滤资源类型为EntityUsingResource且属性4等于您键入的资源的所有Resource4

到目前为止,我已经通过使用弹簧规范使用此标准api来做到这一点:

public static Specification<EntityUsingResource> 
withResource4HavingProperty4Like(String property4) {
    Join<EntityUsingResource, Resource> join = 
        root.join(EntityUsingResource_.resources, JoinType.INNER);
    Join<EntityUsingResource, Resource4> treatedJoin = 
      cb.treat(join, Resource4.class);

    return cb.like(
            treatedJoin.get(Resource4_.property4).get(Property_.value), 
            "%" + property4 + "%");
}

public static Specification<EntityUsingResource> 
withResource2HavingProperty2Like(String property2) {
    Join<EntityUsingResource, Resource> join = 
        root.join(EntityUsingResource_.resources, JoinType.INNER);
    Join<EntityUsingResource, Resource2> treatedJoin = 
        cb.treat(join, Resource2.class);

    return cb.like(
        treatedJoin.get(Resource2_.property2).get(Property_.value),
        "%" + property2 + "%");
}

我将这些规格与弹簧规格实用程序类一起使用,如下所示:哪里(withResource2HavingProperty2Like(property2))。和(withResource4HavingProperty4Like(property4));

然后将其传递给JpaRepository,或多或少将结果返回给gui。

这将在搜索property1时创建以下SQL:

select 
    entity_using_resource0.id as entityId
from 
    entity_using_resource entity_using_resource0
inner join resource resourceas1_ on 
    entity_using_resource0.id=resourceas1_.entity_using_resource_id
inner join resource resourceas2_ on 
    entity_using_resource0.id=resourceas2_.entity_using_resource_id
inner join resource resourceas3_ on 
    entity_using_resource0.id=resourceas3_.entity_using_resource_id
inner join resource resourceas4_ on 
    entity_using_resource0.id=resourceas4_.entity_using_resource_id
where (resourceas4_.property1 like 'property1')
    and (resourceas2_.property1 like '%property1%') limit ...;

问题是,此查询产生大量重复项。我尝试使用distinct来解决问题,但提出了另一个问题。在EntityUsingResource实体中,我有一列为json类型的列,因此无法使用distinct,因为数据库无法比较json值。

[如何编写使用条件api过滤Resource类型的查询?

提前感谢:-)

hibernate jpa spring-data-jpa criteria criteria-api
1个回答
0
投票
如果您的目标是创建有效的查询而不是使用标准API,则可以使用FluentJPA
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