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如何在 Typescript 中定义 Array.reduce 方法?

[英]How can I define Array.reduce method in Typescript?

I have a problem with the reduce method in TypeScript:我对 TypeScript 中的reduce方法有疑问:

const items = {a: 10, b:20, c: 30}
const itemsTotal = Object.keys(items).reduce((accumulator: number, key: keyof typeof items ) => {
    return accumulator + items[key]
  }, 0)

I keep receiving the Typescript error:我不断收到 Typescript 错误:

Argument of type '(accumulator: number, key: "a" | "b" | "c") => number' is not assignable to parameter of type '(previousValue: string, currentValue: string, currentIndex: number, array: string[]) => string'. '(accumulator: number, key: "a" | "b" | "c") => number' 类型的参数不可分配给 '(previousValue: string, currentValue: string, currentIndex: number, array:字符串[]) => 字符串'。

 Types of parameters 'accumulator' and 'previousValue' are incompatible.***

It seems that I need to define the type of the reduce method, but how?看来我需要定义reduce方法的类型,但是怎么做呢?

Object.keys returns a string[] , so you can't apply a reducer function that expects a keyof typeof items to it. Object.keys返回一个string[] ,所以你不能应用一个减速器 function ,它需要一个keyof typeof items

You could use a type assertion, since you know it's valid:您可以使用类型断言,因为您知道它是有效的:

const items = {a: 10, b:20, c: 30};
const itemsTotal = Object.keys(items).reduce((accumulator, key) => {
    return accumulator + items[key as keyof typeof items];
}, 0);

Playground 操场

...but you don't need the key anyway, just use Object.values : ...但无论如何您都不需要密钥,只需使用Object.values

const items = {a: 10, b:20, c: 30};
const itemsTotal = Object.values(items).reduce((accumulator, value) => {
    return accumulator + value;
}, 0);

Playground 操场

(But frankly I'd just use a simple loop .) (但坦率地说,我只是使用一个简单的循环。)

Define your items as a Record of string to number .将您的项目定义为stringnumber记录 It will be clear for the reduce method that the item is a number .对于reduce方法,该项是一个number是很清楚的。

const items: Record<string, number> = {a: 10, b: 20, c: 30}

const itemsTotal = Object.keys(items).reduce((accumulator: number, key: string) => {
    return accumulator + items[key];
}, 0)

You can also skip the braces in the reduce body.您也可以跳过reduce主体中的大括号。

Object.keys(items).reduce((acc: number, key: string) => acc + items[key], 0)

Even more, you can skip the type designations in the reduce because your items are already defined as numbers in the Record .更重要的是,您可以跳过reduce中的类型指定,因为您的items已经在Record中定义为数字。

Object.keys(items).reduce((acc, key) => acc + items[key], 0)

Not the cleanest solution but you can use this snippet:不是最干净的解决方案,但您可以使用此代码段:

const items = { a: 10, b: 20, c: 30 }
const itemsTotal = Object.keys(items).reduce((accumulator, key) => {
    return accumulator + items[key as keyof typeof items]
}, 0)

Key point is to cast key to keyof typeof items关键点是将key转换为keyof typeof items

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