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Typescript自动获取类中的接口属性

typescript

danday74 · 技术社区 · 6 年前

你好打字专家。

我有以下代码,但我必须在类中重复接口属性,否则我会得到:

类错误地实现了接口

Id: number; 其他的属性呢?谢谢

interface INavigation {
  Id: number;
  AppId: number;
  NavId: number;
  Name: string;
  ParentId: string;
  PageURL: string;
  Position: string;
  Active: string;
  Desktop: string;
  Tablet: string;
  Phone: string;
  RoleId: string;
  Target: string;
}

class Navigation implements INavigation {

  Id: number;
  AppId: number;
  NavId: number;
  Name: string;
  ParentId: string;
  PageURL: string;
  Position: string;
  Active: string;
  Desktop: string;
  Tablet: string;
  Phone: string;
  RoleId: string;
  Target: string;

  constructor(navigation: any) {
    this.Id = navigation.Id
    this.AppId = navigation.NavAppId
    this.NavId = navigation.NavId
    this.Name = navigation.NavName
    this.ParentId = navigation.NavParentId
    this.PageURL = navigation.NavPageURL
    this.Position = navigation.NavPosition
    this.Active = navigation.NavActive
    this.Desktop = navigation.NavDesktop
    this.Tablet = navigation.NavTablet
    this.Phone = navigation.NavPhone
    this.RoleId = navigation.NavRoleId
    this.Target = navigation.NavTarget
  }
}

2 回复 | 直到 6 年前

SamB 5 年前

这在Typescript中现在可以使用类/接口合并。

interface Foo {
    a: number;
}

interface Baz extends Foo { }
class Baz {
    constructor() {
        console.log(this.a); // no error here
    }
}

https://github.com/Microsoft/TypeScript/issues/340#issuecomment-184964440

Titian Cernicova-Dragomir 6 年前

但是,我们可以使用返回类的函数作为类的基类型。这个函数可以稍微撒谎,并声称它实现了接口。如果有必要,我们还可以为成员传递一些默认值。

interface INavigation {
  Id: number;
  AppId: number;
  NavId: number;
  Name: string;
  ParentId: string;
  PageURL: string;
  Position: string;
  Active: string;
  Desktop: string;
  Tablet: string;
  Phone: string;
  RoleId: string;
  Target: string;
}

function autoImplement<T>(defaults?: Partial<T>) {
  return class {
    constructor() {
      Object.assign(this, defaults || {});
    }
  } as new () => T
}

class Navigation extends autoImplement<INavigation>() {
  constructor(navigation: any) {
    super();
    // other init here
  }
}

function autoImplementWithBase<TBase extends new (...args: any[]) => any>(base: TBase) {
  return function <T>(defaults?: Partial<T>): Pick<TBase, keyof TBase> & {
    new(...a: (TBase extends new (...o: infer A) => unknown ? A : [])): InstanceType<TBase> & T
  } {
    return class extends base {
      constructor(...a: any[]) {
        super(...a);
        Object.assign(this, defaults || {});
      }
    } as any
  }
}

class BaseClass {
  m() { }
  foo: string
  static staticM() { }
  static staticFoo: string
}

class Navigation extends autoImplementWithBase(BaseClass)<INavigation>() {
  constructor(navigation: any) {
    super();
    // Other init here
  }
}

Navigation.staticFoo
Navigation.staticM
new Navigation(null).m();
new Navigation(null).foo;

Quentin Gary 3 年前

这两个答案之间的混合,通过在构造函数中使用类/接口合并和Object.assign来避免构造函数中的长赋值:

interface Foo {
    a: number;
    b: number;
}

interface Baz extends Foo { }

class Baz  {
    c: number = 4

  constructor (foo: Foo) {
    Object.assign(this, foo, {})
  }
  
  getC() {
    return this.c
  }
}

let foo: Foo = {
    a: 3,
    b: 8
}

let baz = new Baz(foo)
// keep methods and properties
console.warn(baz)

samira mokaram 6 年前

类声明应该显式实现接口。