typescript-interface

interface Girll{
  name:string;
  age:number;  
  bust:number;     
}
const girl={
  name:'大脚',
  age:18,
  bust:100      
}

const screenResume2=( girl:Girl)=>{
  girl.age<24 && girl.bust>=90 && console.log(girl.name+'进入面试') 
  girl.age>24 || girl.bust<90  && console.log(girl.name+'你被淘汰')
}

const getResume2=( girl:Girl)=>{
  console.log(girl.name+'年龄是:'+girl.age)
  console.log(girl.name+'胸围是:'+girl.bust)
}

 

接口非必选定义

interface Girl {
    name : string;
    age  : number;
    bust : number;
    waistline ?: number;
}
const getResume=( girl:Girl)=>{
    console.log(girl.name+'年龄是:'+girl.age)
    console.log(girl.name+'胸围是:'+girl.bust)
    girl.waistline && console.log(girl.name+'腰围是:'+girl.waistline)
}

 

可以写任意参数

interface Girl{
  name:string;
  age:number;
  bust:number;
  waistline?:number;
  [propname:string]:any //这个的意思是,属性的名字是字符串类型,属性的值可以是任何类型。
  say():string  //必须有string的返回值
}

const girl={
  name:'大脚',
  age:18,
  bust:94,
  waistline:21,
  sex:'女',
  say(){
    return '欢迎光临 ,红浪漫洗浴!!'
  }
}

 

接口和类的约束

class xiaojiejie implements Girl{
  name='liuying'
  age=18
  bust=100
  say(){
    return '12312'
  }
}

 

接口间的继承

// 接口间的继承
interface Teacher extends Girl{
  teach():string
}
const girl2={
  name:'大脚',
  age:18,
  bust:94,
  waistline:21,
  sex:'女',
  say(){
      return '欢迎光临 ,红浪漫洗浴!!'
  },
  teach(){
    return '我是一个老师'
  }
}

 

posted on 2020-08-26 15:09  秃了头也不退休  阅读(198)  评论(0编辑  收藏  举报

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