为什么我不能给LiveData分配一个值呢？

Question

我正在学习LiveData，代码A来自https://github.com/googlecodelabs/android-databinding

我认为代码A太复杂了，所以我尝试将其简化为代码B。

但是代码B不能编译，您可以看到图像C，我如何修复它？

码A

class SimpleViewModelSolution : ViewModel() {
    private val _name = MutableLiveData("Ada")
    private val _lastName = MutableLiveData("Lovelace")
    private val _likes =  MutableLiveData(0)

    val name: LiveData<String> = _name
    val lastName: LiveData<String> = _lastName
    val likes: LiveData<Int> = _likes

    // popularity is exposed as LiveData using a Transformation instead of a @Bindable property.
    val popularity: LiveData<Popularity> = Transformations.map(_likes) {
        when {
            it > 9 -> Popularity.STAR
            it > 4 -> Popularity.POPULAR
            else -> Popularity.NORMAL
        }
    }

    fun onLike() {
        _likes.value = (_likes.value ?: 0) + 1
    }
}

码B

class SimpleViewModelSolution : ViewModel() {


    val name: LiveData<String> =MutableLiveData("Ada")
    val lastName: LiveData<String> = MutableLiveData("Lovelace")
    val likes: LiveData<Int> = MutableLiveData(0)

    // popularity is exposed as LiveData using a Transformation instead of a @Bindable property.
    val popularity: LiveData<Popularity> = Transformations.map(likes) {           //The par 'likes' is OK
        when {
            it > 9 -> Popularity.STAR
            it > 4 -> Popularity.POPULAR
            else -> Popularity.NORMAL
        }
    }

    fun onLike() {
        var a=likes.value //It's OK        
        likes.value = (likes.value ?: 0) + 1  //Error
    }
}

图像C

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​

Answer 1

将likes对象初始化为MutableLiveData<>而不是LiveData<>，将代码更改为：

val name: LiveData<String> =MutableLiveData("Ada")
val lastName: LiveData<String> = MutableLiveData("Lovelace")
val likes: MutableLiveData<Int> = MutableLiveData<Int>(0)

// popularity is exposed as LiveData using a Transformation instead of a @Bindable property.
val popularity: LiveData<Popularity> = Transformations.map(likes) {           //The par 'likes' is OK
    when {
        it > 9 -> Popularity.STAR
        it > 4 -> Popularity.POPULAR
        else -> Popularity.NORMAL
    }
}

fun onLike() {
    var a = likes.value //It's OK
    likes.value = (likes.value ?: 0) + 1  //Now it'll work fine
}

Answer 2

您不能直接更改MutableLiveData，LiveData，但是您必须使用名为LiveData的东西，我们可以说LiveData在某种意义上是ImmutableLiveData。MutableLiveData进一步为您提供了两个方法：postValue和setValue，在这里的docs中对这两个方法进行了解释：

https://developer.android.com/reference/android/arch/lifecycle/MutableLiveData

你要做的就是

private val _likes: MutableLiveData<Int> = MutableLiveData()
val likes : LiveData<Int>
get() = _likes

上面我们说过，现在我可以改变LiveData，但是您想提供什么价值，因为您必须这样做

fun changeValue(newValue: Int) {
_likes.value = newValue
}

在你的活动中

vm.likes.observe(this, Observer{
//ui
}

然后任意设置你想要的值

vm.changeValue(1)