Spring Boot Jpa specification Set Enums IN Set Enums

Question

嗨，我想使用Specification按类型过滤餐厅，但我得到了错误：

参数值[SUSHI]与预期类型[java.util.Set(n/a(]不匹配

如何将List enums命令为List enums？：/

我的规范文件：

if(Objects.nonNull(category)){
predicates.add(root.get(Restaurant_.category).in(category));
}

餐厅型号：

@ElementCollection(targetClass = RestaurantCategory.class)
@Enumerated(EnumType.STRING)
@CollectionTable(name = "restaurant_category")
private Set<RestaurantCategory> category;

餐厅类别：

public enum RestaurantCategory {
STREET_FOOD, SUSHI, PIZZA, BURGER
}

Answer 1

遇到了同样的问题，这个问题帮助了我。

就我而言，我想按国家搜索我的佣金。以下是我解决问题的方法。

域：

@Entity
@Table(name = "commission")
data class Commission(
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
var id: Long? = 0,
@ElementCollection(fetch = FetchType.EAGER)
@Enumerated(EnumType.STRING)
var countries: MutableSet<CountryCode>? = null,
@get: NotNull
@Enumerated(EnumType.STRING)
@Column(name = "type", nullable = false)
var type: CommissionType? = null
) 

规格代码(@请参阅findByCountrys部分(：

object CommissionSpecifications {
fun buildQuery(request: SearchCommissionRequest): Specification<Commission> {
var query = Specification.where<Commission>(distinct())
request.type?.also { query = query?.and(findByType(it)) }
request.countries?.also { if (it.isNotEmpty()) query = query?.and(findByCountries(it)) }
return query!!
}
private fun findByType(type: CommissionType): Specification<Commission> =
Specification { root, _, cb -> cb.equal(root.get<Commission>(Commission::type.name), type) }
private fun findByCountries(countries: Set<CountryCode>): Specification<Commission> =
Specification { root, _, _ ->
root.join<Commission, Set<CountryCode>>(Commission::countries.name).`in`(countries)
}
private fun distinct(): Specification<Commission?> {
return Specification { root: Root<Commission?>?, query: CriteriaQuery<*>, cb: CriteriaBuilder? ->
query.distinct(true)
null
}
}

}

Answer 2

我必须执行JOIN。谓词.add(root.join(Restaurant_.category(.in(category((；

同样有效；(