有没有办法合并kotlin数据类而不指定所有属性?
data class MyDataClass(val prop1: String, val prop2: Int, ...//many props)
带有以下签名的函数:
fun merge(left: MyDataClass, right: MyDataClass): MyDataClass
此函数检查两个类上的每个属性以及它们不同的位置使用left参数来创建新的MyDataClass。
这可能使用kotlin-reflect或其他方法吗?
编辑:更清晰
以下是我希望能够做的更好的描述
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"))
}
答案 0 :(得分:1)
如果您希望在左侧的值为null
时从右侧复制值,则可以执行以下操作:
infix inline 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(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)
答案 1 :(得分:0)
您的要求与复制left
值完全相同:
fun merge(left: MyDataClass, right: MyDataClass) = left.copy()
也许使用之一并不能正确理解另一个。请详细说明这不是你想要的。
请注意,由于right
未被使用,您可以将其设为vararg并且&#34;合并&#34;尽你所能:)
fun merge(left: MyDataClass, vararg right: MyDataClass) = left.copy()
val totallyNewData = merge(data1, data2, data3, data4, ...)
修改强>
Kotlin的课程没有跟踪他们的增量。想想你在经历这个过程时得到了什么。第一次更改后,你有
current = Bob("Ben", null, null, "local_id")
next = Bob(null, 1, null, "local_id")
如何知道您希望next
将更改应用于age
而不是name
?如果你只是根据可空性进行更新,
@mfulton有一个很好的答案。否则,您需要自己提供信息。
答案 2 :(得分:0)
当我们可以定义要合并的字段时,组合数据类的特定于类的方法是:
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)
}
}
答案 3 :(得分:0)
@ mfulton26的解决方案仅合并属于主要构造函数的属性。我已将其扩展为支持所有属性
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
}
答案 4 :(得分:-1)
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);
}
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);
};
}
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;