过滤出对象数组并对特定元素进行数学运算答案

【问题标题】：Filtering out array of object and do maths for specific element过滤出对象数组并对特定元素进行数学运算
【发布时间】：2021-08-16 18:28:47
【问题描述】：

我有一个对象数组。比方说

let cartItems = [
   {title: "Veg Salad", category: {title: "Salad"}, quantity: 2, price: 30},
   {title: "Roti", category: {title: "Breads"}, quantity: 3, price: 10},
   {title: "Biriyani", category: {title: "Biriyani"}, quantity: 2, price: 100},
   {title: "Veg Curry", category: {title: "Curry"}, quantity: 3, price: 60},
]

这里 类别 - 沙拉和面包不收取包装费，但 类别 - Biriyani 和咖喱 收取 20 卢比的包装费。所以我想过滤掉其余的数组对象，每个数量增加 20 卢比，并想要总包装费。

const findPackCharge = cartItems.filter((ele) => {
  return (
    ele.category?.title != "Salad" && ele.category?.title != "Breads"
  );
});

console.log(findPackCharge, "FIND");

我尝试了很多，但无法弄清楚。

【问题讨论】：

4个物品中的任何一个都没有包装费。
@Shubanker 上面写的很清楚，沙拉和面包类不收包装费，Biriyani 和咖喱类20卢比包装费
循环过滤后的数组，计算20 * ele.quantity的总数

标签： javascript arrays object filter

【解决方案1】：

不确定您要做什么，但如果只是要获得总包装费，只需将每个项目的数量乘以 20 并保存总和。

const cartItems = [
   {title: "Veg Salad", category: {title: "Salad"}, quantity: 2, price: 30},
   {title: "Roti", category: {title: "Breads"}, quantity: 3, price: 10},
   {title: "Biriyani", category: {title: "Biriyani"}, quantity: 2, price: 100},
   {title: "Veg Curry", category: {title: "Curry"}, quantity: 3, price: 60},
]

const NO_PACKING_CHARGE_ITEMS = ["Breads", "Salad"];
const totalPackingCharge = cartItems.reduce((accu, item) => {
  const isNoPackingChargeItem = NO_PACKING_CHARGE_ITEMS.includes(item.category.title);
  return isNoPackingChargeItem ? accu + (item.quantity * 20) : accu;
}, 0);
console.log(totalPackingCharge);

【讨论】：

【解决方案2】：

使用reduce() 计算总包装费用。在回调函数中可以跳过不收费的项目。

let cartItems = [
   {title: "Veg Salad", category: {title: "Salad"}, quantity: 2, price: 30},
   {title: "Roti", category: {title: "Breads"}, quantity: 3, price: 10},
   {title: "Biriyani", category: {title: "Biriyani"}, quantity: 2, price: 100},
   {title: "Veg Curry", category: {title: "Curry"}, quantity: 3, price: 60},
];

const findPackCharge = cartItems.reduce((total, ele) => {
  if (ele.category?.title != "Salad" && ele.category?.title != "Breads") {
    total += 20 * ele.quantity;
  }
  return total;
}, 0);

console.log(findPackCharge, "FIND");

【讨论】：

【解决方案3】：

const cartItems = [
    { title: "Veg Salad", category: { title: "Salad" }, quantity: 2,  price: 30 },
    { title: "Roti", category: { title: "Breads" }, quantity: 3, price: 10 },
    { title: "Biriyani", category: { title: "Biriyani" }, quantity: 2, price: 100 },
    { title: "Veg Curry", category: { title: "Curry" }, quantity: 3, price: 60 }
];

const packingFee = 20;
const noPackingItems = ["Breads", "Salad"];
const findPackCharge = cartItems.filter((ele) =>         
    !noPackingItems.includes(ele.category?.title)
  ).reduce((total, e) => total + e.quantity * packingFee, 0) ;

console.log(findPackCharge, "for packing charge");

【讨论】：