Typescript noob有关在类构造函数中使用默认道具的参数对象的问题

Question

我有一个要重写为TS的javascript类：

class A {
  constructor({foo: 'foo', bar: 123, baz: true} = {}) {
    this.foo = foo
    this.bar = bar
    this.baz = baz
  }
}

我目前在TS上写的内容：

interface AProps {
  foo: string,
  bar: number,
  baz: boolean,
}

class A {
  foo: string
  bar: number
  baz: boolean

  constructor({foo = 'foo', bar = 123, baz = true}: Partial<AProps> = {}) {
    this.foo = foo
    this.bar = bar
    this.baz = baz
  }
}

如您所见，打字稿的代码多3倍，如何使它更紧凑？据我所知，几乎任何代码重复都是反模式的，因此必须有一种方法告诉打字稿我要从该接口获取类中的所有字段。

拒绝投票时，请至少在评论中写出问题出在哪里

Answer 1

您无需指定AProps接口即可提示构造函数的类型，它将被正确推断。

class A {
  foo: string
  bar: number
  baz: boolean

  constructor({foo = 'foo', bar = 123, baz = true} = {}) {
    this.foo = foo
    this.bar = bar
    this.baz = baz
  }
}

构造函数的类型：

constructor A({ foo, bar, baz }?: {
    foo?: string | undefined;
    bar?: number | undefined;
    baz?: boolean | undefined;
}): A

[const a = new A({x: 1});无法通过消息编译

Argument of type '{ x: number; }' is not assignable to parameter of type '{ foo?: string | undefined; bar?: number | undefined; baz?: boolean | undefined; }'.
  Object literal may only specify known properties, and 'x' does not exist in type '{ foo?: string | undefined; bar?: number | undefined; baz?: boolean | undefined; }'.(2345)

带有原始代码的构造函数的类型为：

constructor A({ foo, bar, baz }?: Partial<AProps>): A

[const a = new A({x: 1});无法通过消息编译

Argument of type '{ x: number; }' is not assignable to parameter of type 'Partial<AProps>'.
  Object literal may only specify known properties, and 'x' does not exist in type 'Partial<AProps>'.(2345)