这个问题直接类似于TypeScript中的类类型检查

我需要在运行时找出任何类型的变量是否实现了接口。这是我的代码:

interface A{
    member:string;
}

var a:any={member:"foobar"};

if(a instanceof A) alert(a.member);

如果您在typescript游乐场中输入这段代码,最后一行将被标记为错误,“名称A不存在于当前作用域”。但事实并非如此,该名称确实存在于当前作用域中。我甚至可以更改变量声明为var a: a ={成员:"foobar"};没有编辑的抱怨。在浏览网页并找到其他问题后,我将接口更改为类,但我不能使用对象字面量来创建实例。

我想知道A类型是如何消失的,但看看生成的javascript就能解释这个问题:

var a = {
    member: "foobar"
};
if(a instanceof A) {
    alert(a.member);
}

没有将A表示为接口,因此不可能进行运行时类型检查。

我知道javascript作为一种动态语言没有接口的概念。是否有方法对接口进行类型检查?

typescript游乐场的自动完成显示typescript甚至提供了一个方法实现。我怎么使用它?


当前回答

基于Fenton的回答,下面是我实现的一个函数,用于验证给定对象是否具有接口所具有的键(全部或部分)。

根据您的用例,您可能还需要检查每个接口属性的类型。下面的代码不会这样做。

function implementsTKeys<T>(obj: any, keys: (keyof T)[]): obj is T {
    if (!obj || !Array.isArray(keys)) {
        return false;
    }

    const implementKeys = keys.reduce((impl, key) => impl && key in obj, true);

    return implementKeys;
}

用法示例:

interface A {
    propOfA: string;
    methodOfA: Function;
}

let objectA: any = { propOfA: '' };

// Check if objectA partially implements A
let implementsA = implementsTKeys<A>(objectA, ['propOfA']);

console.log(implementsA); // true

objectA.methodOfA = () => true;

// Check if objectA fully implements A
implementsA = implementsTKeys<A>(objectA, ['propOfA', 'methodOfA']);

console.log(implementsA); // true

objectA = {};

// Check again if objectA fully implements A
implementsA = implementsTKeys<A>(objectA, ['propOfA', 'methodOfA']);

console.log(implementsA); // false, as objectA now is an empty object

其他回答

我在filter-descriptor.interface.d.ts文件中的@progress/kendo-data-query中找到了一个例子

检查程序

declare const isCompositeFilterDescriptor: (source: FilterDescriptor | CompositeFilterDescriptor) => source is CompositeFilterDescriptor;

示例使用

const filters: Array<FilterDescriptor | CompositeFilterDescriptor> = filter.filters;

filters.forEach((element: FilterDescriptor | CompositeFilterDescriptor) => {
    if (isCompositeFilterDescriptor(element)) {
        // element type is CompositeFilterDescriptor
    } else {
        // element type is FilterDescriptor
    }
});

下面是我使用类和lodash想出的解决方案:(它有效!)

// TypeChecks.ts
import _ from 'lodash';

export class BakedChecker {
    private map: Map<string, string>;

    public constructor(keys: string[], types: string[]) {
        this.map = new Map<string, string>(keys.map((k, i) => {
            return [k, types[i]];
        }));
        if (this.map.has('__optional'))
            this.map.delete('__optional');
    }

    getBakedKeys() : string[] {
        return Array.from(this.map.keys());
    }

    getBakedType(key: string) : string {
        return this.map.has(key) ? this.map.get(key) : "notfound";
    }
}

export interface ICheckerTemplate {
    __optional?: any;
    [propName: string]: any;
}

export function bakeChecker(template : ICheckerTemplate) : BakedChecker {
    let keys = _.keysIn(template);
    if ('__optional' in template) {
        keys = keys.concat(_.keysIn(template.__optional).map(k => '?' + k));
    }
    return new BakedChecker(keys, keys.map(k => {
        const path = k.startsWith('?') ? '__optional.' + k.substr(1) : k;
        const val = _.get(template, path);
        if (typeof val === 'object') return val;
        return typeof val;
    }));
}

export default function checkType<T>(obj: any, template: BakedChecker) : obj is T {
    const o_keys = _.keysIn(obj);
    const t_keys = _.difference(template.getBakedKeys(), ['__optional']);
    return t_keys.every(tk => {
        if (tk.startsWith('?')) {
            const ak = tk.substr(1);
            if (o_keys.includes(ak)) {
                const tt = template.getBakedType(tk);
                if (typeof tt === 'string')
                    return typeof _.get(obj, ak) === tt;
                else {
                    return checkType<any>(_.get(obj, ak), tt);
                }
            }
            return true;
        }
        else {
            if (o_keys.includes(tk)) {
                const tt = template.getBakedType(tk);
                if (typeof tt === 'string')
                    return typeof _.get(obj, tk) === tt;
                else {
                    return checkType<any>(_.get(obj, tk), tt);
                }
            }
            return false;
        }
    });
}

自定义类:

// MyClasses.ts

import checkType, { bakeChecker } from './TypeChecks';

class Foo {
    a?: string;
    b: boolean;
    c: number;

    public static _checker = bakeChecker({
        __optional: {
            a: ""
        },
        b: false,
        c: 0
    });
}

class Bar {
    my_string?: string;
    another_string: string;
    foo?: Foo;

    public static _checker = bakeChecker({
        __optional: {
            my_string: "",
            foo: Foo._checker
        },
        another_string: ""
    });
}

在运行时检查类型:

if (checkType<Bar>(foreign_object, Bar._checker)) { ... }

与上面使用用户定义守卫的情况相同,但这次使用的是箭头函数谓词

interface A {
  member:string;
}

const check = (p: any): p is A => p.hasOwnProperty('member');

var foo: any = { member: "foobar" };
if (check(foo))
    alert(foo.member);

Typescript 2.0引入了带标签的联合

Typescript 2.0特性

interface Square {
    kind: "square";
    size: number;
}

interface Rectangle {
    kind: "rectangle";
    width: number;
    height: number;
}

interface Circle {
    kind: "circle";
    radius: number;
}

type Shape = Square | Rectangle | Circle;

function area(s: Shape) {
    // In the following switch statement, the type of s is narrowed in each case clause
    // according to the value of the discriminant property, thus allowing the other properties
    // of that variant to be accessed without a type assertion.
    switch (s.kind) {
        case "square": return s.size * s.size;
        case "rectangle": return s.width * s.height;
        case "circle": return Math.PI * s.radius * s.radius;
    }
}

Typescript中的类型保护:

TS有用于此目的的类型保护。他们是这样定义的:

执行运行时检查以保证类型的表达式 在某种范围内。

这基本上意味着TS编译器在拥有足够的信息时可以将类型缩小到更特定的类型。例如:

function foo (arg: number | string) {
    if (typeof arg === 'number') {
        // fine, type number has toFixed method
        arg.toFixed()
    } else {
        // Property 'toFixed' does not exist on type 'string'. Did you mean 'fixed'?
        arg.toFixed()
        // TSC can infer that the type is string because 
        // the possibility of type number is eliminated at the if statement
    }
}

回到您的问题,我们还可以将类型保护的概念应用于对象,以确定它们的类型。要为对象定义类型保护,需要定义一个返回类型为类型谓词的函数。例如:

interface Dog {
    bark: () => void;
}

// The function isDog is a user defined type guard
// the return type: 'pet is Dog' is a type predicate, 
// it determines whether the object is a Dog
function isDog(pet: object): pet is Dog {
  return (pet as Dog).bark !== undefined;
}

const dog: any = {bark: () => {console.log('woof')}};

if (isDog(dog)) {
    // TS now knows that objects within this if statement are always type Dog
    // This is because the type guard isDog narrowed down the type to Dog
    dog.bark();
}