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  • typescript 技巧学习

    typescript 差缺补漏

    资料

    交叉点类型

    交集类型是将多个类型组合为一种的方法,就是多个类型的合并

    type LeftType = {
      id: number
      left: string
    }
    
    type RightType = {
      id: number
      right: string
    }
    
    type IntersectionType = LeftType & RightType
    
    function showType(args: IntersectionType) {
      console.log(args)
    }
    
    showType({ id: 1, left: "test", right: "test" })
    // Output: {id: 1, left: "test", right: "test"}
    

    联合类型用的比较多

    let a:number|string;
    

    通用类型感觉这个很厉害

    构造泛型类型,用方括号T作为参数传递

    interface GenericType<T> {
      id: number,
      name: T
    }
    
    function showType(args: GenericType<string>) {
      console.log(args);
    }
    
    function showTypeTwo(args: GenericType<number>) {
      console.log(args);
    }
    
    showType({id: 10, name: 'xxx'});
    showTypeTwo({id: 10, name: 20});
    

    我以前以为只能用T,今天看了大佬的代码,发现原来T只是一个符号,学习到了

    interface GenericType<T,D>{
      id:T,
      name:D
    }
    
    function showType(args: GenericType<number, string>) {
      console.log(args);
    }
    showType({id:20, name: 'ss'});
    
    function showTypeTwo(args: GenericType<number, Array<string>>) {
      
    }
    showTypeTwo({id:10, name: ['a', 'b', 'c']})
    

    可选

    Partial<T>

    interface PartialType {
    id: number
    firstName: string
    lastName: string
    }
    
    function showType(args: Partial<PartialType>) {
    console.log(args)
    }
    
    showType({ id: 1 })
    // Output: {id: 1}
    
    showType({ firstName: "John", lastName: "Doe" })
    // Output: {firstName: "John", lastName: "Doe"}
    

    必传

    Required<T>

    提供的T需要传所有类型的属性

    interface RequiredType {
    id: number
    firstName?: string
    lastName?: string
    }
    
    function showType(args: Required<RequiredType>) {
    console.log(args)
    }
    
    showType({ id: 1, firstName: "John", lastName: "Doe" })
    // Output: { id: 1, firstName: "John", lastName: "Doe" }
    
    showType({ id: 1 })
    //报错啦
    

    只读

    Readonly<T>

    T无法使用新值重新分配

    interface ReadonlyType {
    id: number
    name: string
    }
    
    function showType(args: Readonly<ReadonlyType>) {
    args.id = 4
    console.log(args)
    }
    showType({ id: 1, name: "Doe" })
    // error 报错了,他的值都不能被修改
    不过下面这样写也不错
    interface ReadonlyType {
    readonly id: number
    name: string
    }
    

    忽略

    Omit<T,k>

    将K从属性T 删除

    简单的说就是我输入类型不能带入k,不然会报错

    interface PickType {
    id: number
    firstName: string
    lastName: string
    }
    
    function showType(args: Omit<PickType, "firstName" | "lastName">) {
    console.log(args)
    }
    showType({id: 12})
    // 正常
    // 带有firstName 或者 lastName就会报错
    

    挑选

    Pick<T,K>

    T中要有属性k

    interface PickType {
    id: number
    firstName: string
    lastName: string,
    name: string
    }
    
    function showType(args: Pick<PickType, 'firstName' | 'lastName'>) {
    console.log(args);
    }
    
    showType({lastName: 'xx', firstName: 'xx'});
    showType({lastName: 'xx', firstName: 'xx', id: 12});
    // 报错
    showType({lastName: 'xx', firstName: 'xx', name:'xx'});
    // 报错
    

    仔细的理解了下,就是类型中只能包含firstName 和 lastName,所以可以理解成挑选

    k就是你要选择的属性,你可以用|进行分隔选择多个字段

    记录

    Record<K,T>

    将一个类型的属性K映射到另一个类型的属性

    interface EmployeeType {
    id: number
    fullname: string
    role: string
    }
    
    let employees: Record<number, EmployeeType> = {
    0: { id: 1, fullname: "John Doe", role: "Designer" },
    1: { id: 2, fullname: "Ibrahima Fall", role: "Developer" },
    2: { id: 3, fullname: "Sara Duckson", role: "Developer" },
    }
    // 0: { id: 1, fullname: "John Doe", role: "Designer" },
    // 1: { id: 2, fullname: "Ibrahima Fall", role: "Developer" },
    // 2: { id: 3, fullname: "Sara Duckson", role: "Developer" }
    

    映射类型

    把每个属性转换为新类型

    type StringMap<T> = {
    [P in keyof T]: string
    }
    
    function showType(arg: StringMap<{ id: number; name: string }>) {
    console.log(arg)
    }
    showType({name:'xx',id:'ss'})
    //就是把已经的属性都转成string
    showType({ id: 1, name: "Test" })
    // 会报错
    

    类型防护

    typeof

    function showType(x: number | string) {
      if (typeof x === "number") {
        return `The result is ${x + x}`
      }
      throw new Error(`This operation can't be done on a ${typeof x}`)
    }
    
    showType("I'm not a number")
    // Error: This operation can't be done on a string
    
    showType(7)
    // Output: The result is 14
    

    instanceof

    class Foo {
      bar() {
        return "Hello World"
      }
    }
    
    class Bar {
      baz = "123"
    }
    
    function showType(arg: Foo | Bar) {
      if (arg instanceof Foo) {
        console.log(arg.bar())
        return arg.bar()
      }
    
      throw new Error("The type is not supported")
    }
    
    showType(new Foo())
    // Output: Hello World
    
    showType(new Bar())
    // Error: The type is not supported
    

    in

    忽略的点看来我之前的写法没这种好

    interface Bookes {
      id: number,
      name: string
    }
    
    books: Bookes[] = [
        {id: 1, name: 'Angular'},
        {id: 2, name: 'Typescript'},
        {id: 3, name: 'Javascript'},
        {id: 4, name: 'HTML'}
      ];
    

    发现一个奇怪的!

    !只是用来判断null和undefined;

    as则可用于变更(缩小或者放大都可以)类型检测范围

    as和!用于属性的读取,都可以缩小类型检查范围,都做判空用途时是等价的。只是!具体用于告知编译器此值不可能为空值(null和undefined),而as不限于此

    ?可用于属性的定义和读取,读取时告诉编译器此值可能为空值(null和undefined),需要做判断。

    interface InterArray {
      children?: InterArrayOne
    }
    
    interface InterArrayOne {
      name: string,
      age: number
    }
    export class OneComponent implements OnInit {
      public a: (InterArrayOne & InterArray) [] = [{
        name: 'xx',
        age: 12,
        children: {
          name: 'aaa',
          age: 12
        }
      },
        {
          name: 'aa',
          age: 13,
        },
    
      ];
      add(): void {
        for (let i = 0; i < this.a.length; i++) {
          console.log(this.a[i]!.children?.name);
        }
      }
      ngOnInit(): void {
        this.add();
      }
    }
    

    类型谓词

    function isString(x: any): x is string {
      return typeof x === 'string';
    }
    

    决定自己的高度的是你的态度,而不是你的才能

    记得我们是终身初学者和学习者

    总有一天我也能成为大佬

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  • 原文地址:https://www.cnblogs.com/fangdongdemao/p/14307005.html
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