通过ID比较2个不同的数组并计算差异 [英] Compare 2 different Arrays by ID and calculate difference

问题描述

我有2个数组

ArrayA = <代码> {数据":{"PlayerList":[{平台":1，"PlayerExternalId":205288，价格":250，剩余时间":22}，{"Platform":1，"PlayerExternalId":205753，"Price":10000，"RemainingTime":22}]}}

ArrayB = <代码> {"datafut":[{"currentPricePs4":"4149000"，"currentPriceXbox":"3328000"，"PlayerExternalId":"151152967"}}，{"currentPricePs4:" 3315000，" currentPriceXbox:" 2720000，" PlayerExternalId:" 151198320}}}

ArrayB就像一个比较价格的小型数据库.从理论上讲，ArrayA需要通过ArrayB 拦截.但这会创建一个新的ArrayC，这对我来说很复杂，因为我需要ArrayA的结果索引.

ArrayB is like a small database to compare prices. ArrayA needs theoretically an Interception with ArrayB. But this creates a new ArrayC which is complicated for me because I need the index of the results from ArrayA.

此外，在比较两个数组ID时，我需要比较两个价格并将差异计算为变量，以便以后使用.我该如何实现?

Moreover when comparing both array IDs, I need to compare both prices and calculate a difference into a variable so I can work with it later. How can I achieve this?

这是我的伪代码.如果这是正确的方法，那就知道.

This is my pseudo code. Idk if this is even the right way..

Filter ArrayB by ArrayA //by playerID
for(
NewPrice = ArrayA.price / ArrayB.price + Index of ArrayA.price
index = Index of ArrayA.price)

或者我可以将价格从arrayB追加到arrayA并可以通过某种方式进行计算吗?

or could I append the price from arrayB to arrayA and can calculate then somehow?

推荐答案

您可以将两个数组都传递给以下函数:我已经存储了索引，现在如果只需要索引，则不需要对其进行排序，否则我将进行排序它以索引为基础来保持原始顺序.

You can pass both arrays to following function: I have stored index, now if you only need index, you don't need to sort it otherwise I am sorting it on the base of index to keep the original order.

function mergeArrays(arrayA, arrayB) {

    var players = arrayA.data.PlayerList;
    var data = arrayB.data;
    var arrayC = [];

    for(let i=0; i<data.length; i++) {  
        var playerId = data[i].PlayerExternalId;
        for(let j=0; j<players.length; j++) {
            if(players[j].PlayerExternalId != playerId) {
                continue;
            }
            var obj = {};
            obj.playerId = playerId;
            obj.index = j;
            obj.price = players[j].price;
            obj.xboxprice = data[i].currentPriceXbox;
            obj.ps4price = data[i].currentPricePs4;
            arrayC.push(obj);
        }
    }
    arrayC.sort((a,b) => (a.index < b.index)?-1:(a.index>b.index?1:0));
    return arrayC;
}

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