TypeScript：仅对特殊类公开字段

Question

这是一个问题。我使用模式MVVM，这意味着我有一些仅具有显示逻辑的视图 和一些仅具有功能逻辑的ViewModels。 ViewModels 有方法和字段。并且这些字段和方法可以被其他ViewModel或者ViewModel的View读取或调用。

我想知道，有没有办法让某些字段仅在 View 中可见，但在其他 ViewModels 中隐藏（私有）？

我需要这样的东西：

class ViewModel1 {
  doSomething = () => {...}; // can be called anywhere

  publicForView doSomethingInView = () => {...}; // can be called only in the View

  private doSomethingInViewModel = () => {...}; // can be called only here
}

// view is a HOC-function, which joins ViewModel with View
const View1 = view(ViewModel1, ({ viewModel }) => {
  viewModel.doSomethingInView(); // can be called
  viewModel.doSomething(); // can be called
  viewModel.doSomethingInViewModel(); // can't be called

  return <div/>;
});

class ViewModel2 {
  private viewModel1: ViewModel1;

  constructor() {
    viewModel.doSomethingInView(); // can't be called
    viewModel.doSomething(); // can be called
    viewModel.doSomethingInViewModel(); // can't be called
  }
}

Answer 1

如果您希望 Typescript 中的类仅公开类的某些部分，您可以创建接口，因此使用 Pick / Omit 节省重复代码将有助于创建接口。

例如..

class ViewModel {
  doSomething = () => {};  
  doSomethingInView = () => {}
  doSomethingInViewModel = () => {}
}

type View = Omit<ViewModel, 'doSomethingInViewModel'>
type Model = Omit<ViewModel, 'doSomethingInView'>

function doSomethingWithView(view: View) {
    view.doSomething();
    view.doSomethingInView();
    //view.doSomethingInViewModel();  will error
}

function doSomethingWidthModel(model: Model) {
    model.doSomething();
    model.doSomethingInViewModel();
    // view.doSomethingInView(); will error
}

const view = new ViewModel();
//we can pass view to both functions, but the methods will be restricted
doSomethingWithView(view);
doSomethingWidthModel(view);

这里的工作示例 -> TS Playground