我发现在DTO中使用继承的最好方法是使用Lombok插件在java中创建数据类。

不要忘记在注释中将lombok.equalsAndHashCode.callSuper设置为true

可以从非数据类继承数据类。

基类

open class BaseEntity (

@ColumnInfo(name = "name") var name: String? = null,
@ColumnInfo(name = "description") var description: String? = null,
// ...
)

子类

@Entity(tableName = "items", indices = [Index(value = ["item_id"])])
data class CustomEntity(

    @PrimaryKey
    @ColumnInfo(name = "id") var id: Long? = null,
    @ColumnInfo(name = "item_id") var itemId: Long = 0,
    @ColumnInfo(name = "item_color") var color: Int? = null

) : BaseEntity()

它工作。

可以从非数据类继承数据类。不允许从另一个数据类继承数据类，因为在继承的情况下，无法使编译器生成的数据类方法一致且直观地工作。

Kotlin Traits可以提供帮助。

interface IBase {
    val prop:String
}

interface IDerived : IBase {
    val derived_prop:String
}

数据类

data class Base(override val prop:String) : IBase

data class Derived(override val derived_prop:String,
                   private val base:IBase) :  IDerived, IBase by base

示例使用

val b = Base("base")
val d = Derived("derived", b)

print(d.prop) //prints "base", accessing base class property
print(d.derived_prop) //prints "derived"

这种方法也可以用来解决@Parcelize的继承问题

@Parcelize 
data class Base(override val prop:Any) : IBase, Parcelable

@Parcelize // works fine
data class Derived(override val derived_prop:Any,
                   private val base:IBase) : IBase by base, IDerived, Parcelable

在构造函数外的超类中将属性声明为抽象，并在子类中重写它们。

abstract class Resource {
    abstract var id: Long
    abstract var location: String
}

data class Book (
    override var id: Long = 0,
    override var location: String = "",
    var isbn: String
) : Resource()

虽然在层次结构中正确地实现equals()确实相当困难，但支持继承其他方法仍然很好，例如:toString()。

为了更具体一点，让我们假设我们有以下结构(显然，它不起作用，因为toString()不是继承的，但如果它不是很好吗?):

abstract class ResourceId(open val basePath: BasePath, open val id: Id) {

    // non of the subtypes inherit this... unfortunately...
    override fun toString(): String = "/${basePath.value}/${id.value}"
}

data class UserResourceId(override val id: UserId) : ResourceId(UserBasePath, id)

data class LocationResourceId(override val id: LocationId) : ResourceId(LocationBasePath, id)

Assuming our User and Location entities return their appropriate resource IDs (UserResourceId and LocationResourceId respectively), calling toString() on any ResourceId could result in quite a nice little representation that is generally valid for all subtypes: /users/4587, /locations/23, etc. Unfortunately, because non of the subtypes inherited to overridden toString() method from the abstract base ResourceId, calling toString() actually results in a less pretty representation: <UserResourceId(id=UserId(value=4587))>, <LocationResourceId(id=LocationId(value=23))>

还有其他方法可以对上述模型进行建模，但这些方法要么迫使我们使用非数据类(错过了数据类的许多好处)，要么我们最终在所有数据类中复制/重复toString()实现(没有继承)。