用TypeScript中的逻辑扩展接口？[重复]

i5desfxk 于 2023-02-13 发布在 TypeScript

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typescript default function inside interface（3个答案）
昨天关门了。
我有一个Interface，我想在其中创建使用其他属性的计算属性。
例如，我的Person接口有名字和姓氏属性，我如何扩展Person接口以提供一个名为fullName的新属性供所有实现者组合名字和姓氏属性？

interface Person {
  firstName: string;
  lastName: string;
}

class Pilot implements Person {
     constructor(public firstName: string, public lastName: string) {}
}

class Sailer implements Person {
     constructor(public firstName: string, public lastName: string) {}
}

const pilot = new Pilot("Joe", "Alpha")
const sailer = new Sailer("Jane", "Beta")

// Extend `Person` interface to return firstName + lastName?

console.log(pilot.fullName) // Joe Alpha
console.log(sailer.fullName) // Jane Beta

typescript

来源：https://stackoverflow.com/questions/75421276/extend-interfaces-with-logic-in-typescript

3条答案

按热度按时间

oalqel3c1#

对于这种情况，我将继续使用父类

interface IPerson {
  firstName: string;
  lastName: string;
}

class Person implements IPerson {
  constructor(public firstName: string, public lastName: string) {}

  getFullName() {
    return this.firstName + ' ' + this.lastName;
  }
}

class Pilot extends Person {}

class Sailer extends Person {}

const pilot = new Pilot("Joe", "Alpha")
const sailer = new Sailer("Jane", "Beta")

console.log(pilot.getFullName())
console.log(sailer.getFullName())

这是fiddle。

赞(0）回复(0）举报 2023-02-13

webghufk2#

扩展Person接口以添加fullName属性的一种方法是使用getter：

interface Person {
  firstName: string;
  lastName: string;
  fullName: string;
}

class Pilot implements Person {
  firstName: string;
  lastName: string;

  constructor(firstName: string, lastName: string) {
    this.firstName = firstName;
    this.lastName = lastName;
  }

  get fullName(): string {
    return `${this.firstName} ${this.lastName}`;
  }
}

class Sailor implements Person {
  firstName: string;
  lastName: string;

  constructor(firstName: string, lastName: string) {
    this.firstName = firstName;
    this.lastName = lastName;
  }

  get fullName(): string {
    return `${this.firstName} ${this.lastName}`;
  }
}

const pilot = new Pilot("Joe", "Alpha");
const sailor = new Sailor("Jane", "Beta");

console.log(pilot.fullName); // Joe Alpha
console.log(sailor.fullName); // Jane Beta

赞(0）回复(0）举报 2023-02-13

zte4gxcn3#

我认为在TypeScript中将getter函数绑定到接口是不可能的，但是你可以使用一个父类Person来实现一个接口IPerson，来达到你想要的效果：

interface IPerson {
  firstName: string;
  lastName: string;
}

class Person implements IPerson {
  constructor(public firstName: string, public lastName: string) {}
  public get fullName() {
    return `${this.firstName} ${this.lastName}`;
  }
}

class Pilot extends Person {
  constructor(public firstName: string, public lastName: string) {
    super(firstName, lastName);
  }
}

class Sailer extends Person {
  constructor(public firstName: string, public lastName: string) {
    super(firstName, lastName);
  }
}

const pilot = new Pilot("Joe", "Alpha");
const sailer = new Sailer("Jane", "Beta");

console.log(pilot.fullName); // Joe Alpha
console.log(sailer.fullName); // Jane Beta

赞(0）回复(0）举报 2023-02-13

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用TypeScript中的逻辑扩展接口？[重复]

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