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

问题描述

在TypeScript 1.7中使用 Polymorphic ，因为我发现这里，我们可以在一个返回类中定义一个方法类型的 this ，并且自动地，扩展该类并继承这些方法的任何类将其返回类型设置为它们各自的 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，我要做的是继承 static 方法引用类本身的返回类型。最好在代码中描述：

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?)
}

也许我得考虑一下一个不同的方式来实现它，因为TypeScript 1.7中的Polymorphic this 不支持 static 方法 / 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：我猜我们会看到这个Github问题是如何包含的：

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T中是可行的ypeScript 2.0+。通过使用内嵌的 {new（）：T} 类型来捕获这个，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:

class BaseModel {

  static getAll<T>(this: { new(): T }): T[] {
    return [] // dummy impl
  }

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

class SubModel extends BaseModel {
}

const sub = new SubModel()
const savedSub: SubModel = sub.save()
const savedSubs: SubModel[] = SubModel.getAll()

请注意， getAll 仍然没有参数。

Note that getAll still expects no arguments with this typing.

有关更多信息，请参见 https://www.typescriptlang.org/docs/handbook/generics.html#泛型使用类类型和 https://stackoverflow.com/a/45262288/126801 6

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

这篇关于TypeScript：在继承类中为静态方法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！