4

有没有办法在不指定所有属性的情况下合并 kotlin 数据类？

data class MyDataClass(val prop1: String, val prop2: Int, ...//many props)

具有以下签名的函数：

fun merge(left: MyDataClass, right: MyDataClass): MyDataClass

此函数检查两个类的每个属性以及它们不同的位置，使用左侧参数创建一个新的 MyDataClass。

这可能使用 kotlin-reflect 或其他方式吗？

编辑：更清晰

这是对我想要做的事情的更好描述

  data class Bob(
        val name: String?,
        val age: Int?,
        val remoteId: String?,
        val id: String)

@Test
fun bob(){

    val original = Bob(id = "local_id", name = null, age = null, remoteId = null)
    val withName = original.copy(name = "Ben")
    val withAge = original.copy(age = 1)
    val withRemoteId = original.copy(remoteId = "remote_id")

    //TODO: merge without accessing all properties
    // val result = 
    assertThat(result).isEqualTo(Bob(id = "local_id", name = "Ben", age=1, remoteId = "remote_id"))
}
4

5 回答 5

13

如果要在左侧的值是左侧的值时从右侧复制值，null则可以执行以下操作：

inline infix fun <reified T : Any> T.merge(other: T): T {
    val propertiesByName = T::class.declaredMemberProperties.associateBy { it.name }
    val primaryConstructor = T::class.primaryConstructor
        ?: throw IllegalArgumentException("merge type must have a primary constructor")
    val args = primaryConstructor.parameters.associateWith { parameter ->
        val property = propertiesByName[parameter.name]
            ?: throw IllegalStateException("no declared member property found with name '${parameter.name}'")
        (property.get(this) ?: property.get(other))
    }
    return primaryConstructor.callBy(args)
}

用法：

data class MyDataClass(val prop1: String?, val prop2: Int?)
val a = MyDataClass(null, 1)
val b = MyDataClass("b", 2)
val c = a merge b // MyDataClass(prop1=b, prop2=1)
于 2017-06-15T14:53:32.350 回答
2

当我们可以定义要组合的字段时，一种特定于类的组合数据类的方法是：

data class SomeData(val dataA: Int?, val dataB: String?, val dataC: Boolean?) {
    fun combine(newData: SomeData): SomeData {        
        //Let values of new data replace corresponding values of this instance, otherwise fall back on the current values.
        return this.copy(dataA = newData.dataA ?: dataA,
                dataB = newData.dataB ?: dataB,
                dataC = newData.dataC ?: dataC)
    }
}
于 2018-04-02T05:37:23.817 回答
1

@mfulton26 的解决方案仅合并属于主构造函数的属性。我已经扩展它以支持所有属性

inline infix fun <reified T : Any> T.merge(other: T): T {
    val nameToProperty = T::class.declaredMemberProperties.associateBy { it.name }
    val primaryConstructor = T::class.primaryConstructor!!
    val args = primaryConstructor.parameters.associate { parameter ->
        val property = nameToProperty[parameter.name]!!
        parameter to (property.get(other) ?: property.get(this))
    }
    val mergedObject = primaryConstructor.callBy(args)
    nameToProperty.values.forEach { it ->
        run {
            val property = it as KMutableProperty<*>
            val value = property.javaGetter!!.invoke(other) ?: property.javaGetter!!.invoke(this)
            property.javaSetter!!.invoke(mergedObject, value)
        }
    }
    return mergedObject
}
于 2018-05-15T06:50:11.773 回答
0

left您的要求与复制值完全相同：

fun merge(left: MyDataClass, right: MyDataClass) = left.copy()

也许其中一种用途没有正确理解另一种。如果这不是您想要的，请详细说明。

请注意，由于right未使用，因此您可以将其设为可变参数并“合并”任意数量:)

fun merge(left: MyDataClass, vararg right: MyDataClass) = left.copy()

val totallyNewData = merge(data1, data2, data3, data4, ...)

编辑

Kotlin 中的类不跟踪它们的增量。想想你在这个过程中得到了什么。第一次更改后，您有

current = Bob("Ben", null, null, "local_id")
next = Bob(null, 1, null, "local_id")

它应该如何知道您想要next将更改应用于age但不是name？如果您只是基于可空性进行更新，@mfulton 有一个很好的答案。否则，您需要自己提供信息。

于 2017-06-15T14:50:24.187 回答
-2 
infix fun <T : Any> T.merge(mapping: KProperty1<T, *>.() -> Any?): T {
    //data class always has primary constructor ---v
    val constructor = this::class.primaryConstructor!!
    //calculate the property order
    val order = constructor.parameters.mapIndexed { index, it -> it.name to index }
                                      .associate { it };

    // merge properties
    @Suppress("UNCHECKED_CAST")
    val merged = (this::class as KClass<T>).declaredMemberProperties
                                           .sortedWith(compareBy{ order[it.name]})
                                           .map { it.mapping() }
                                           .toTypedArray()


    return constructor.call(*merged);
}

编辑

infix fun <T : Any> T.merge(right: T): T {
    val left = this;
    return left merge mapping@ {
        //    v--- implement your own merge strategy
        return@mapping this.get(left) ?: this.get(right);
    };
}

例子

val original = Bob(id = "local_id", name = null, age = null, remoteId = null)
val withName = original.copy(name = "Ben")
val withAge = original.copy(age = 1)
val withRemoteId = original.copy(remoteId = "remote_id")

val result = withName merge withAge merge withRemoteId;
于 2017-06-15T11:54:07.760 回答