java.lang.RuntimeException：无法创建类 ViewModel 的实例

Question

我正在尝试实例化一个 viewModel 类，该类包含一个 init 块，我正在尝试通过 volley 发送 Get 请求。问题是当我试图在我的 Fragment 类中实例化这个类时，我收到了这个错误，我不知道为什么。也许来自构造函数的上下文会是问题？谢谢：

ViewModel 类

class BooksFragmentViewModel(c : Context) : ViewModel() {

lateinit var cont: Context
var books: MutableLiveData<MutableList<BookItem>> = MutableLiveData<MutableList<BookItem>>()
var triggerAddBook = MutableLiveData<Boolean>()
var triggerDeleteBook = MutableLiveData<Boolean>()
var triggerEditBook = MutableLiveData<Boolean>()
var bookRepo: BookRepository = BookRepository()
var bookItemAdd: BookItem? = null
var bookItemDelete: BookItem? = null
var bookItemEdit : BookItem? = null
var bookNew: LiveData<BookItem> = Transformations.switchMap(triggerAddBook) {
    if (it != null && it)
        addBook()
    else
        null
}

  init {
        books = bookRepo.booksGetRequest(c)
    }

片段类

class BooksFragment : Fragment(), BooksAdapter.OnDeleteBtnClicked, BooksAdapter.OnEditBtnClicked {
lateinit var booksModel: BooksFragmentViewModel
var position: Int = 0

private var adapter: BooksAdapter = BooksAdapter(this, this)
override fun onCreateView(inflater: LayoutInflater, container: ViewGroup?,
                          savedInstanceState: Bundle?): View? {
    return inflater.inflate(R.layout.fragment_books, container, false)
}

override fun onViewCreated(view: View, savedInstanceState: Bundle?) {
    booksModel = ViewModelProvider(requireActivity()).get(BooksFragmentViewModel::class.java)
    booksLoading_progressBar.visibility = View.VISIBLE
    var c: Context? = context
    if (c != null) {
        booksModel.setContext(c)
    }

Answer 1

问题确实是因为带有参数的主构造函数。您要求ViewModelProvider 创建BooksFragmentViewModel 类的实例，但提供者对预期参数一无所知。

Here is a post 关于该主题。

如何解决这个问题？

1. 尽可能去掉所有主要的构造函数参数。这应该是最简单的方法；
2. 创建ViewModelProvider.Factory，它将处理您的视图模型的创建。

第一个选项很清楚，但是对于第二个选项，您应该执行以下操作。

创建ViewModelProvider.Factory:

class BooksFragmentViewModelFactory (val context: Context) :
    ViewModelProvider.Factory {

    override fun <T : ViewModel?> create(modelClass: Class<T>): T {
        if (modelClass.isAssignableFrom(BooksFragmentViewModel::class.java)) {
            return BooksFragmentViewModel(context) as T
        } else {
            throw RuntimeException("Failed to create instance of the ${modelClass.simpleName}")
        }
    }
}

使用工厂ViewModelProvider:

    override fun onViewCreated(view: View, savedInstanceState: Bundle?) {
        val bookViewModelFactory = BooksFragmentViewModelFactory(requireContext())
        booksModel = ViewModelProvider(requireActivity(), bookViewModelFactory).get(BooksFragmentViewModel::class.java)
        booksLoading_progressBar.visibility = View.VISIBLE
    }

更多解释请查看codelab from Google (step 8)。有足够的示例来了解如何使用视图模型工厂的基础知识。

Answer 2

如果您有带参数的ViewModel（如示例中的上下文），那么您需要使用Factory 将任何参数传递给ViewModel

 booksModel = ViewModelProvider(this,object :
        ViewModelProvider.Factory{
        override fun <T : ViewModel?> create(modelClass: Class<T>): T {
            return booksModel(requireActivity()) as T
        }
    }).get(booksModel::class.java)