1

我有一个对象:

class User {
  var id: String? = null
  var name: String? = null
}

和对列表:

val fieldsToChange = listOf<Pair<String, String>>(Pair("name", "foo"), Pair("id", "bar"))

我想迭代槽对列表并使用反射为给定属性设置适当的值。

4

2 回答 2

2

给定类实例obj,我们可以使用obj::class.memberProperties.

我们可以构造一个从属性名称到属性的映射:

val nameToProperty = obj::class.memberProperties.associateBy(KProperty<*>::name)

然后我们可以迭代fieldsToChange并检索属性并设置它:

fieldsToChange.forEach { (propertyName, propertyValue) ->
   nameToProperty[propertyName]
       .takeIf { it is KMutableProperty<*> } // take only "settable" (var) properties
       ?.let { it as KMutableProperty<*> } // cast it to mutable property so we can access setter
       ?.let { it.setter.call(obj, propertyValue) } // call the setter
}

此外,我们可以使其通用:

fun setFields(obj: Any, fieldsToChange: List<Pair<String, Any?>>) {
    val nameToProperty = obj::class.memberProperties.associateBy(KProperty<*>::name)
    fieldsToChange.forEach { (propertyName, propertyValue) ->
        nameToProperty[propertyName]
                .takeIf { it is KMutableProperty<*> }
                ?.let { it as KMutableProperty<*> }
                ?.let { it.setter.call(obj, propertyValue) }
    }
}

val user = User()
setFields(user, fieldsToChange)

assert(user.name == "foo")
assert(user.id == "bar")

可能的改进是优化nameToProperty以仅包含已转换为 KMutableProperty 的 MutableProperties

于 2019-07-18T09:28:05.097 回答
2

您可以使用地图作为代表:

class User(map: Map<String, String>) {
    val id: String by map
    val name: String by map
}

val fieldsToChange = listOf(Pair("name", "foo"), Pair("id", "bar"))
val map = fieldsToChange.map { it.first to it.second }.toMap()

val user = User(map)
于 2019-07-18T15:36:57.650 回答