javascript - typescript :放大或缩小接收对象？

Question

我的应用程序中有一些食谱，我想将它们放大或缩小到卡路里量。

我的接收对象看起来像这样:

{
  title: "some title",
  calories: 700,
  fat: 20,
  carbs: 10,
  protein: 40,
  //...
  ingredient: [
    {
      name: "sugar",
      amount: 20,
      measurement: "g"
    },
    {
      name: "banana",
      amount: 1,
      measurement: "piece"
    }
  ]
}

我的问题:将热量增加到 1000 卡路里或减少到例如 400 卡路里的最佳方法是什么？

是否可以用整个对象做一个经典的“3 组”:首先计算此食谱的 100 卡路里，然后计算所有属性的 400、500 或 1000 卡路里？

感谢您的帮助:)

Answer 1

这似乎有效:

interface Ingredient {
    name: string;
    amount: number;
    measurement: "g" | "piece";
}

interface Reciepe {
    title: string;
    calories: number;
    fat: number;
    carbs: number;
    protein: number;
    ingredient: Ingredient[];
}

function scale(reciepe: Reciepe, calories: number): Reciepe {
    const ratio = calories / reciepe.calories;
    const scaled = Object.assign({}, reciepe);

    scaled.calories = calories;
    scaled.fat = reciepe.fat * ratio;
    scaled.carbs = reciepe.carbs * ratio;
    scaled.protein = reciepe.protein * ratio;

    for (let i = 0; i < reciepe.ingredient.length; i++) {
        scaled.ingredient[i].amount = reciepe.ingredient[i].amount * ratio;
    }

    return scaled;
}

( code in playground )