根据Typescript中的输入键从具有动态键的对象中选择值

Pick value from object with dynamic key based on input key in Typescript

所以我的后端有一个像这样的JSON

name: "Hello",
nameTranslations: {nl:"Hallo", en: "Hello", fr: "Bonjour"}
description: "Hello",
descriptionTranslations: {nl:"Hallo", en: "Hello", fr: "Bonjour"}

我想构建一个函数,我将在其中放入对象和键,它会自动找到可能的最佳翻译。在 Javascript 中很容易,但我想正确使用 Typescript,这样我就不会在忘记查询翻译时处理运行时错误(我使用 GraphQL)

function getText(object, key, language) {
  const best = object[`${key}Translations`][language];
  if (best) {
    return best;
  }

  return object[key];
}

我有一些 Typescript 要求

更新 根据@michmich112 的回答,我确实使用了以下函数

function translate<T extends BackendTranslations>(translations: T, fallback: string, language: keyof BackendTranslations) => {
      return translations[language] || fallback;
}

translate(project.nameTranslations, project.name)

我会保留答案,因为我仍然对这是否可以在 Typescript 中工作感兴趣

首先,您需要为 JSON 创建一个界面:

interface ITranslationData {
  nl: string
  en: string
  fr: string
}

interface ITranslation {
  name: string
  nameTranslations: ITranslationData
  description: string
  descriptionTranslations: ITranslationData
}

完成后,您可以按如下方式在函数中使用类型:

function getText(obj: ITranslation, 
  key: keyof ITranslation,
  language: keyof ITranslationData): string {
  const best = obj[key][language];
  if (best) {
    return best;
  }

  return obj[key];
}

这将要求您的密钥准确无误,这意味着您将无法执行 const best = object[${key}Translations][language];

这也意味着您可以在一行中编写代码:

function getText(obj: ITranslation, 
  key: keyof ITranslation,
  language: keyof ITranslationData): string {
  return obj[key][language] || obj[key];
}

这是我的尝试:

const json = {
  name: "Hello",
  nameTranslations: { nl: "Hallo", en: "Hello", fr: "Bonjour" },
  description: "Hello",
  descriptionTranslations: { nl: "Hallo", en: "Hello", fr: "Bonjour" }
}

type BackendTranslations = {
  nl: string;
  en: string;
  fr: string;
}


type Raw<T> = T extends `${infer R}Translations` ? R : never

/**
 * Once you will have more specific interface for json
 * this overloading will help you,
 * for now it always returns string , because all properties/nested properties of
 * JSON are strings
 */
//function getText<K extends Raw<keyof Data>>(obj: Data, key: K, language: keyof Translations): K extends string ? Data[`${K}Translations`][keyof Translations] : Data[K];

function getText<
  Data extends Record<string, unknown>,
  K extends Raw<keyof Data>
>(obj: Data, key: K, language: keyof BackendTranslations) {
  return (
    (obj[`${key}Translations`] as BackendTranslations)[language] || obj[key]
  );
}

const result = getText(json, 'name', 'en') // ok
const result1 = getText(json, 'nameX', 'en') // error
const result2 = getText(json, 'nameTranslations', 'en') // error
const result3 = getText(json, 'description', 'en') // ok

请记住,我的解决方案仅适用于 TypeScript 4。*