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javascript - 循环遍历一个对象中的多个数组以在一个新数组中创建多个对象？

问题描述

在我的应用程序中，$scope.results = [{},{},{},{}.....]。（是一个包含多个对象的数组）

对象之一的示例输出：

0:
brand: "Confidential"
group_id: "CRCLEBESP"
id: "d703187ac59976b066c9b7ea01416bca"
msrp: "0"
name: "Clear Beyond Speaker Cable"
price: "5510"
product_type_unigram: "cable"
sku: "CRCLEBESP"
ss_msrp: (12) ["0", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0"]
ss_msrp_max: "0"
ss_msrp_min: "0"
ss_name_sizes: "Cardas Clear Beyond Speaker Cable"
ss_price: (12) ["6840", "5510", "9500", "8170", "8170", "12160", "10830", "14820", "13490", "17480", "16150", "18810"]
ss_price_max: "18810"
ss_price_min: "5510"
stock_id: "171038"
sub_sku: (12) ["1.5PSPSP", "1PSPSP", "2.5PSPSP", "2PBNBN", "2PSPSP", "3.5PSPSP", "3PSPSP", "4.5PSPSP", "4PSPSP", "5.5PSPSP", "5PSPSP", "6PSPSP"]
uid: "171038"
__proto__: Object

这些对象包含 3 个数组，每个数组我需要从名为“ss_msrp、ss_price 和 sub_sku”中提取值。每个对象中的每个数组都是相同的长度。例如，对象 0 有 3 个数组，每个数组有 12 个索引长。对象 1 有 3 个数组，每个数组有 6 个索引长。对象 3 有 3 个数组，每个数组有 8 个索引等。

我需要从每个数组中提取第 [n] 个项目并将其放入一个新对象中。每个对象将有 3 个键值对。例如：

ss_msrp = ["0", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0"]
ss_price = ["6840", "5510", "9500", "8170", "8170", "12160", "10830", "14820", "13490", "17480", "16150", "18810"]
sub_sku = ["1.5PSPSP", "1PSPSP", "2.5PSPSP", "2PBNBN", "2PSPSP", "3.5PSPSP", "3PSPSP", "4.5PSPSP", "4PSPSP", "5.5PSPSP", "5PSPSP", "6PSPSP"]

object1 = { msrp: "0", price: "6840", sku: "1.5PSPSP" }
object2 = { msrp: "0", price: "5510", sku: "1PSPSP" }
object3 = { msrp: "0", price: "9500", sku: "2.5PSPSP" }
object4 = { msrp: "0", price: "8170", sku: "2PBNBN" }

我的代码现在看起来像这样：

let objects = []

for (let i=0; i < $scope.results[0].sub_sku.length; i++){   
    objects.push({
        sku: $scope.results[0].sub_sku[i],
        msrp: $scope.results[0].ss_msrp[i],
        price: $scope.results[0].ss_price[i]
        })
    }

这当然只会返回第一个对象的所需结果。我曾尝试将其嵌套在另一个 for 循环中，如下所示：

for (let x=0; x < $scope.results.length; x++){
                for (let i=0; i < $scope.results[x].sub_sku.length; i++){   
                    objects.push({
                        sku: $scope.results[x].sub_sku[i],
                        msrp: $scope.results[x].ss_msrp[i],
                        price: $scope.results[x].ss_price[i]
                    })
                }
            }

但我收到以下错误： 无法读取未定义的属性“长度”

有人有想法么？先感谢您。

标签： javascriptarrays

解决方案

假设您的输入在数组中有 2 个对象

例子：

let examples=[
    {
    brand: "Confidential",
    group_id: "CRCLEBESP",
    id: "d703187ac59976b066csds9b7ea01416bca",
    msrp: "0",
    name: "Clear Beyond Speaker Cable",
    price: "5510",
    product_type_unigram: "cable",
    sku: "CRCLEBESP",
    ss_msrp:["0", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0"],
    ss_msrp_max: "0",
    ss_msrp_min: "0",
    ss_name_sizes: "Cardas Clear Beyond Speaker Cable",
    ss_price:["6840", "5510", "9500", "8170", "8170", "12160", "10830", "14820", "13490", "17480", "16150", "18810"],
    ss_price_max: "18810",
    ss_price_min: "5510",
    stock_id: "171038",
    sub_sku:["1.5PSPSP", "1PSPSP", "2.5PSPSP", "2PBNBN", "2PSPSP", "3.5PSPSP", "3PSPSP", "4.5PSPSP", "4PSPSP", "5.5PSPSP", "5PSPSP", "6PSPSP"],
    uid: "171038",
},
{
    brand: "Confidential",
    group_id: "CRCLEBESP",
    id: "d703187ac59976b066c9b7ea01416bca",
    msrp: "0",
    name: "Clear Beyond Speaker Cable",
    price: "5510",
    product_type_unigram: "cable",
    sku: "CRCLEBESP",
    ss_msrp:["0", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0", "0"],
    ss_msrp_max: "0",
    ss_msrp_min: "0",
    ss_name_sizes: "Cardas Clear Beyond Speaker Cable",
    ss_price:["6840", "5510", "9500", "8170", "8170", "12160", "10830", "14820", "13490", "17480", "16150", "18810"],
    ss_price_max: "18810",
    ss_price_min: "5510",
    stock_id: "171038",
    sub_sku:["1.5PSPSP", "1PSPSP", "2.5PSPSP", "2PBNBN", "2PSPSP", "3.5PSPSP", "3PSPSP", "4.5PSPSP", "4PSPSP", "5.5PSPSP", "5PSPSP", "6PSPSP"],
    uid: "171038",
}]

您可以处理2个案例，

  1. 如果您的对象数组包含2 个相同的产品，并且您需要 2 个重复的对象，其中包含price、sku和msrp ，那么这是解决方案之一。
let result = examples.map(row=>{
   return row.ss_msrp.map((e,i)=>{
        return {
            msrp:row.ss_msrp[i],
            sku:row.sub_sku[i],
            price:row.ss_price[i]
        };
   });
});

Output:
// console.log(result);
// (2) [Array(12), Array(12)]
// It will print an array which contains array of objects of each product
// eg: [[{msrp: "0", sku: "5PSPSP", price: "16150"},....(12 entries)],[{msrp: "0", sku: "5PSPSP", price: "16150"},....(12 entries)]]

2.如果您不希望重复的产品也引用产品 id，如果 id 是唯一的，那么您也可以这样做。

let result = {};
examples.map(row=>{
    result[row.id]= row.ss_msrp.map((e,i)=>{
        return {
        msrp:row.ss_msrp[i],
        sku:row.sub_sku[i],
        price:row.ss_price[i]
        };
    });
});

Output:
//console.log(result);
// {d703187ac59976b066c9b7ea01416bca: (12) [{…}, {…}, {…}, {…}, {…}, {…}, {…}, {…}, {…}, {…}, {…}, {…}]}

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