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，因此我尝试将其简化为代码B 。

但是代码B 无法编译，您可以看到图像C ，我该如何解决？

代码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
    }
}

代码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
    }
}

图片C

Answer 1

将likes对象初始化为MutableLiveData <>而不是LiveData <> 将您的代码更改为：

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 LiveData ，但必须使用 MutableLiveData ，我们可以说 LiveData 为 ImmutableLiveData 。 MutableLiveData 进一步为您提供了另外两种方法 postValue 和 setValue ，在文档中对此进行了说明：
https://developer.android.com/reference/android/arch/lifecycle/MutableLiveData

您要做的一切

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

我们上面说过，现在我可以更改LiveData，但是您想提供什么价值，因为您要做的是

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

您的活动内

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

然后根据需要设置值

vm.changeValue(1)

为什么不能为LiveData分配值？

2 个答案: