使用强类型地图 [英] Using strong typed Map

问题描述

我很难用typescript 1.8.10键入Map对象.以下是core-js定义Map接口的摘录:

I'm having trouble strong typing my Map objects with typescript 1.8.10. Here is an excerpt from core-js defining the Map interface:

interface Map<K, V> {
    clear(): void;
    delete(key: K): boolean;
    forEach(callbackfn: (value: V, index: K, map: Map<K, V>) => void, thisArg?: any): void;
    get(key: K): V;
    has(key: K): boolean;
    set(key: K, value?: V): Map<K, V>;
    size: number;
}

我想创建一个使用字符串键的映射，并且只存储形状为{name:string,price:number}的值.我尝试用以下方法声明对象:

I want to create a map that uses string keys and only ever stores values with the shape {name:string,price:number}. I tried declaring my object with:

let oMap:Map<string,{name:string,price:number}> = new Map();

但是，编译器会引发错误TS2322: Type 'Map<{}, {}>' is not assignable to type 'Map<string, { name: string; price: number; }>'.在打字稿中使用ES6 Map对象时，有没有办法利用强打字?

However, the compiler throws error TS2322: Type 'Map<{}, {}>' is not assignable to type 'Map<string, { name: string; price: number; }>'. Is there a way to take advantage of strong typing when using ES6 Map objects in typescript?

推荐答案

您需要像这样向创建的Map提供通用类型信息:

You need to provide generic types information to the created Map like that:

let oMap:Map<string,{name:string,price:number}> = new Map<string,{name:string,price:number}>();

然后，您可以省略类型声明，而将工作留给编译器:

And after that you can omit type declaration, leaving the job to compiler:

// oMap is correctly inferred to be Map<string,{name:string,price:number}>
let oMap = new Map<string,{name:string,price:number}>();

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