枚举列类型上的Hibernate@filter

Question

@Entity
@Table(name = "table_1")
@TypeDefs(
  TypeDef(name = "enum", typeClass = PostgreSQLEnumType::class)
)
@Mockable
class table1{

  @Id
  @GeneratedValue(strategy = GenerationType.IDENTITY)
  var id: Int = 0

  ...

  @Enumerated(EnumType.STRING)
  @Type(type = "enum")
  lateinit var enumColumn: EnumClass
}

使用EnumClass：

enum class EnumClass{
  TYPEA(EnumSubClass.ONE),
  TYPEB(EnumSubClass.TWO),
  ...
  TYPEN(EnumSubClass.N);

  val category: EnumSubClass

  constructor(category: EnumSubClass) {
    this.category = category
  }

  companion object {
    ...
  }
}

筛选器位于父实体上：

@Entity
@Table(name = "mla_simulation_turbine")
@FilterDefs(
  FilterDef(name = "enumTypeFilter", parameters = [ParamDef(name="enumTypeParam", type="string")])
)
@Mockable
class ParentEntity{

  @Id
  @GeneratedValue(strategy = GenerationType.IDENTITY)
  var id: Int = 0


  @OneToMany(mappedBy = "...", fetch = FetchType.LAZY)
  @Filters(
    Filter(name = "enumTypeFilter", condition="enumColumn= :enumTypeParam")
  )
  var components = mutableSetOf<EnumClass>()


}

过滤器是这样设置的：

val existingSession = em.unwrap(Session::class.java)
      existingSession.enableFilter("enumTypeFilter")
        .setParameter("enumTypeParam", EnumClass.TYPEA.toString())

SELECT ...
where
        enum_column.enum_type= ? 

Caused by: org.postgresql.util.PSQLException: ERROR: operator does not exist: database_enum= character varying
  Hint: No operator matches the given name and argument type(s). You might need to add explicit type casts.
  Position: 925

和其他几个。

这里的任何帮助或提示都将不胜感激。谢了！

Answer 1

好的，为任何可以从中受益的人找到了解决方案。

您需要将列强制转换为文本，而不是将参数强制转换为枚举....

cast(enumColumn as text) in (:enumTypeParam)

像魅力一样有效。