Red_Baron
Red_Baron

Reputation: 150

Compare 2 different Arrays by ID and calculate difference

I got 2 arrays

ArrayA = {"data":{"PlayerList":[{"Platform":1,"PlayerExternalId":205288,"Price":250,"RemainingTime":22},{"Platform":1,"PlayerExternalId":205753,"Price":10000,"RemainingTime":22}]}}

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

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.

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?

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)

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

Upvotes: 0

Views: 46

Answers (1)

Hamza Arshad
Hamza Arshad

Reputation: 846

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;
}

Upvotes: 1

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