我有一个抽象类,其中有一个MediatorLiveData
对象。该对象具有许多源,其中之一取决于子类,并且在父类中为abstract
。
在init
块中添加源会在运行时导致NullPointerException,因为在init块添加源时,它仍然是抽象的(或者使我相信)。
是否有一种方法可以使用abstract
LiveData
作为MediatorLiveData
的源,而不必在子类中设置该源?我只想override val
并完成它,因为我肯定会在将来的某个时候忘记调用addSources()
函数。
(我知道此示例不是完成此操作的最有用方法,但我不想增加不必要的复杂性)
示例:
abstract class MyClass: ViewModel(){
private val _myMediator = MediatorLiveData<String>()
protected abstract val mySource: LiveData<String>
val myObservable: LiveData<String>
get() = _myMediator
// This will cause a NullPointerException at runtime
init{
_myMediator.addSource(mySource){ _myMediator.value = it }
}
//This should work, but requires this to be called in child class
protected fun addSources(){
_myMediator.addSource(mySource){ _myMediator.value = it }
}
}
class myChild: MyClass(){
override val mySource = Transformations.map(myRepository.someData) { it.toString() }
// This is where init { addSources() } would be called
}
在阅读了Stachu的答案后,我决定使用这个,虽然我没有测试,但是我认为应该可以:
abstract class MyFixedClass: ViewModel(){
private val _myMediator: MediatorLiveData<String> by lazy{
MediatorLiveData<String>().apply{
addSource(mySource){ this.value = it }
}
}
protected abstract val mySource: LiveData<String>
val myObservable: LiveData<String>
get() = _myMediator
}
class MyChild: MyFixedClass(){
override val mySource = Transformations.map(myRepository.someData) { it.toString() }
}
答案 0 :(得分:2)
如何使用惰性评估,例如像这样的东西
abstract class MyClass : ViewModel() {
private val _myMediator = MediatorLiveData<String>()
private val _mySource: LiveData<String> by lazy { mySource() }
protected abstract fun mySource(): LiveData<String>
val myObservable: LiveData<String>
get() = _myMediator
init {
_myMediator.addSource(_mySource) { _myMediator.value = it }
}
}
class myChild : MyClass() {
override fun mySource() = Transformations.map(myRepository.someData) { it.toString() }
}