我正在尝试使用Java中的某种“ COALESCE”函数来连接Javascript中的多个数组。
下面是我的数组:
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
我的第二列是价格
我希望价格列首先从“ 特价商品”数组中检索价格,然后让没有值的行从我的“ mainprices ”返回价格数组。
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);
此输出:
[{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"}]
我现在要做的是将空值替换为“ mainprices.price ”值
如果有人可以帮助我,我将不胜感激!
欢呼
答案 0 :(得分:1)
您可以.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)
而不是O(N^2)
复杂度,将specials
首先映射到其itemid
:
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);
答案 1 :(得分:1)
您可以通过建立itemid
至price
的字典来做到这一点。要构建字典,请在同一数组中列出价格,最后以最重要的价格列出,这样在构建字典时它们会覆盖以前的条目。
从该词典中,可以轻松获取itemid
的价格,并且可以通过映射商品来生成输出。
您可以将此逻辑包装在可重用的函数中,该函数可以使用所需数量的价格数组。您只需要按重要性输入这些数组,最重要的显示在右边。
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);