COALESCE加入JAVASCRIPT

Question

我试图在Javascript中使用某种“COALESCE”函数加入几个数组，就像在SQL中一样。

以下是我的数组：

var items = [
    {itemid: 'A'},
    {itemid: 'B'},
    {itemid: 'C'},
    {itemid: 'D'},
    {itemid: 'E'}
];

var specials = [
    {itemid: 'A', price: '5.00'},
    {itemid: 'C', price: '5.00'},
    {itemid: 'E', price: '5.00'}
];

var mainprices = [
    {itemid: 'A', price: '10.00'},
    {itemid: 'B', price: '10.00'},
    {itemid: 'C', price: '10.00'},
    {itemid: 'D', price: '10.00'},
    {itemid: 'E', price: '10.00'}
];

我想要的是：我的第一列是itemid 我的第二栏是价格

我希望price列首先从我的'specials'数组中检索价格，然后是没有值的行从我的'mainprices'数组返回价格。

这是我到目前为止所尝试的：

var results = [];
for (var i=0; i<items.length; i++) {
    var found = false;
    for (var j=0; j<specials.length; j++) {
        if (items[i].itemid === specials[j].itemid) {
            results.push({
                item_id: items[i].itemid, 
                price_value: specials[j].price
            });
            found = true;
            break;
        }
    }
    if (found === false) {
        results.push({
            item_id: items[i].itemid, 
            price_value: null
        });    
    }
}

console.log(results);

这输出：

[{item_id: "A", price_value: "5.00"}, {item_id: "B", price_value: null}, {item_id: "C", price_value: "5.00"}, {item_id: "D", price_value: null}, {item_id: "E", price_value: "5.00"}]

我现在要做的是用'mainprices.price'值替换空值

如果有人能帮助我，我真的很感激！

干杯

Answer 1

你可以.map mainPrices数组，并在specials中搜索匹配的元素：

var items = [
    {itemid: 'A'},
    {itemid: 'B'},
    {itemid: 'C'},
    {itemid: 'D'},
    {itemid: 'E'}
];

var specials = [
    {itemid: 'A', price: '5.00'},
    {itemid: 'C', price: '5.00'},
    {itemid: 'E', price: '5.00'}
];

var mainprices = [
    {itemid: 'A', price: '10.00'},
    {itemid: 'B', price: '10.00'},
    {itemid: 'C', price: '10.00'},
    {itemid: 'D', price: '10.00'},
    {itemid: 'E', price: '10.00'}
];

const pricesWithSpecials = mainprices.map(({ itemid, price }) => {
  const found = specials.find((e) => e.itemid === itemid);
  return { itemid, price: found ? found.price : price };
});
console.log(pricesWithSpecials);

或者，对于O(N)复杂性而不是O(N^2)复杂性，首先将specials映射到他们的itemids：

var items = [
    {itemid: 'A'},
    {itemid: 'B'},
    {itemid: 'C'},
    {itemid: 'D'},
    {itemid: 'E'}
];

var specials = [
    {itemid: 'A', price: '5.00'},
    {itemid: 'C', price: '5.00'},
    {itemid: 'E', price: '5.00'}
];

var mainprices = [
    {itemid: 'A', price: '10.00'},
    {itemid: 'B', price: '10.00'},
    {itemid: 'C', price: '10.00'},
    {itemid: 'D', price: '10.00'},
    {itemid: 'E', price: '10.00'}
];

const specialsByItemid = specials.reduce((a, item) => {
  a[item.itemid] = item;
  return a;
}, {});

const pricesWithSpecials = mainprices.map(({ itemid, price }) => {
  return { itemid, price: specialsByItemid[itemid] ? specialsByItemid[itemid].price : price };
});
console.log(pricesWithSpecials);

Answer 2

你可以通过建立itemid到price的字典来做到这一点。要构建字典，请在同一个数组中列出价格，最后列出最重要的价格，以便在构建字典时覆盖以前的条目。

从这本词典中，很容易得到你的itemid的价格，你可以通过映射你的项目来生成你的输出。

您可以将此逻辑包装在可重用的函数中，该函数可以根据需要使用尽可能多的价格数组。您只需按重要性提供这些数组，最重要的是右侧。

const items = [
  {itemid: 'A'},
  {itemid: 'B'},
  {itemid: 'C'},
  {itemid: 'D'},
  {itemid: 'E'}
];

const specials = [
  {itemid: 'A', price: '5.00'},
  {itemid: 'C', price: '5.00'},
  {itemid: 'E', price: '5.00'}
];

const mainprices = [
  {itemid: 'A', price: '10.00'},
  {itemid: 'B', price: '10.00'},
  {itemid: 'C', price: '10.00'},
  {itemid: 'D', price: '10.00'},
  {itemid: 'E', price: '10.00'}
];

const superImportantPrices = [
  {itemid: 'B', price: '15.00'}
];

const coalesce = (items, ...prices) => {
  const pricesDict = prices.flat().reduce((acc, { itemid, price }) => {
    acc[itemid] = price;
    return acc;
  }, {});
  return items.map(({ itemid }) => ({ itemid, price: pricesDict[itemid] }));
}

// call coalesce with the most important prices to the right
const result = coalesce(items, mainprices, specials, superImportantPrices);

console.log(result);