在相应的片段中初始化数据绑定变量

Question

我一直在学习 developer.android.com 上的 Android Kotlin 基础课程，并且运行遇到了数据绑定问题。 project I am working on doesn't have solution code provided, but I have been modeling my approach off of a previous similar Codelab.

我正在尝试初始化布局 xml 文件中与每个布局相对应的片段中声明的数据绑定变量，但是当我尝试初始化片段变量时，我收到错误消息：“分类器 'EntreeMenuFragment'没有同伴 object，因此必须在这里初始化”。 build.Gradle dataBinding 和 viewBinding 都设置为 true。

<layout
    ...
    
    <data>
        <variable
            name="viewModel"
            type="com.example.lunchtray.model.OrderViewModel" />

        <variable
            name="EntreeMenuFragment"
            type="com.example.lunchtray.ui.order.EntreeMenuFragment" />
    </data>
   
   ...
</layout>

class EntreeMenuFragment : Fragment() {
    private var _binding: FragmentEntreeMenuBinding? = null
    private val binding get() = _binding!!
    private val sharedViewModel: OrderViewModel by activityViewModels()

    override fun onCreate(savedInstanceState: Bundle?) {
        super.onCreate(savedInstanceState)
    }

    override fun onCreateView(
        inflater: LayoutInflater, container: ViewGroup?,
        savedInstanceState: Bundle?
    ): View? {
        _binding = FragmentEntreeMenuBinding.inflate(inflater, container, false)
        val root: View = binding.root
        return root
    }

    override fun onViewCreated(view: View, savedInstanceState: Bundle?) {
        super.onViewCreated(view, savedInstanceState)
        binding.apply {
            lifecycleOwner = viewLifecycleOwner
            viewModel = sharedViewModel
            EntreeMenuFragment = this@EntreeMenuFragment  // ERROR
        }
    }

    ....

Answer 1

删除绑定变量的大写即可避免错误：

<variable
            name="entreeMenuFragment"
            type="com.example.lunchtray.ui.order.EntreeMenuFragment" />

            entreeMenuFragment = this@EntreeMenuFragment  // fixed

在相应的片段中初始化数据绑定变量

Initializing data binding variable in corresponding fragment

android

kotlin

android-databinding

android-viewmodel

android-viewbinding