TypeScript:继承类中静态方法的自引用返回类型 [英] TypeScript: self-referencing return type for static methods in inheriting classes

问题描述

使用 Polymorphic this 在 TypeScript 1.7 中，正如我发现的那样 这里，我们可以在有返回类型的类中定义方法this 的，自动地，任何扩展该类并继承这些方法的类，都会将它们的返回类型设置为各自的 this 类型.像这样:

With Polymorphic this in TypeScript 1.7, as I discovered here, we can define a method in a class with a return type of this, and automatically, any classes that extend that class and inherit the methods, will have their return types set to their respective this type. Like so:

class Model {
  save():this {    // return type: Model
    // save the current instance and return it
  }
}

class SomeModel extends Model {
  // inherits the save() method - return type: SomeModel
}

然而，我所追求的是拥有一个继承的 static 方法，其返回类型引用类本身.最好用代码来描述:

However, what I'm after is to have an inherited static method with a return type referencing the class itself. It's best described in code:

class Model {
  static getAll():Model[] {
    // return all recorded instances of Model as an array
  }

  save():this {
    // save the current instance and return it
  }
}

class SomeModel extends Model {
  // inherits the save() method - return type: SomeModel
  // also inherits getAll() - return type: Model (how can we make that SomeModel?)
}

也许我必须想出一种不同的方式来实现这一点，因为 TypeScript 1.7 中的多态 this 不支持 static 方法设计em>.

Perhaps I'll have to think of a different way to implement this, since Polymorphic this in TypeScript 1.7 does not support static methods by design.

编辑:我想我们会看到这个 Github 问题是如何结束的:https://github.com/Microsoft/TypeScript/issues/5863

EDIT: I guess we'll see how this Github issue wraps up: https://github.com/Microsoft/TypeScript/issues/5863

推荐答案

这在 TypeScript 2.0+ 中是可行的.通过使用内联 { new(): T } 类型来捕获 this，你会得到你想要的:

This is doable in TypeScript 2.0+. By using an inline { new(): T } type to capture this, you'll get what you wanted:

type Constructor<T> = { new (): T }

class BaseModel {
  static getAll<T>(this: Constructor<T>): T[] {
    return [] // dummy impl
  }
  
  /**
   * Example of static method with an argument:
   */
  static getById<T>(this: Constructor<T>, id: number): T | undefined {
    return // dummy impl
  }

  save(): this {
    return this // dummy impl
  }
}

class SubModel extends BaseModel {}

const sub = new SubModel()
const savedSub: SubModel = sub.save()

// Behold: SubModel.getAll() returns SubModels, not BaseModel
const savedSubs: SubModel[] = SubModel.getAll()

请注意，getAll 仍然不希望使用这种类型的参数.

Note that getAll still expects no arguments with this typing.

有关更多信息，请参阅https://www.typescriptlang.org/docs/handbook/2/generics.html#using-class-types-in-generics 和 https://stackoverflow.com/a/45262288/1268016

For more information, see https://www.typescriptlang.org/docs/handbook/2/generics.html#using-class-types-in-generics and https://stackoverflow.com/a/45262288/1268016

这篇关于TypeScript:继承类中静态方法的自引用返回类型的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！