2

I have a list of products that I wanna sort. Sorting by price and duration is an easy one. But I wanna find the BEST overall producte as well.

Here is what I have so far:

   var divList = $(".listing-item");

    /* -------------------------------------- */

    $("#btnPrice").click(function () {
       divList.sort(function(a, b) { 
         return $(a).data("price")-$(b).data("price")
       });
       $("#list").append(divList);
    });
    
    /* -------------------------------------- */

    $("#btnDuration").click(function () {
       divList.sort(function(a, b) { 
         return $(a).data("duration")-$(b).data("duration")
       });
       $("#list").append(divList);
    });

    /* -------------------------------------- */
    
    $("#btnBest").click(function () {
       /* TBD */
    });
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div id="list">
    <div class="listing-item" data-price="20" data-duration="400" data-stops="3">Price 20, Duration 400, Stops 3</div>
    <div class="listing-item" data-price="30" data-duration="350" data-stops="3">Price 30, Duration 350, Stops 3</div>
    <div class="listing-item" data-price="10" data-duration="700" data-stops="2">Price 10, Duration 700, Stops 2</div>
    <div class="listing-item" data-price="40" data-duration="500" data-stops="1">Price 40, Duration 500, Stops 1</div>
</div>

    <button id="btnPrice" type="button">Sort by price</button>
    <button id="btnDuration" type="button">Sort by duration</button>
    <button id="btnBest" type="button">Sort by overall best</button>

The BEST overall product should have a cheap price, little duration and as less stops as possible.

I am thinking about kind of a "Matrix Calculation" here, comparing each factor of each product with the same attribute of all the other products. At the end we'll have a sum of all factor results.

e.g.

Product 1 has price of 20 and duration of 400 and stops = 3.
Product 2 has price of 30 and duration of 350 and stops = 3.

I would calculate as follows:

A

20/30 = 0.66
400/350 = 1.14
3/3 = 1
SUM = 1.8

B

30/20 = 1.5
350/400 = 0.875
3/3 = 1
SUM = 2.375

In this Calculation Product A wins, because the sum of all factors is less than the sum of all factors for product B.

So far for the theory. But how can I achive this with JS??

Supun Abesekara
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JonSnow
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  • **(1)** How do you calculate your sum? **(2)** What's the meaning of dividing each product's properties with the other? **(3)** How are you going to compare more than 2 products? – GalAbra Feb 22 '18 at 18:30
  • 1+2+3: I don't have a solution yet. For a) I think I will store results in an array and save it in another data-attribute then, so I can sort my list by the sum result of each product. For b): my way of finding a best factor. Smallest sum = best product. For c) have a loop about each product I guess – JonSnow Feb 22 '18 at 22:36
  • are these answers to the questions I’ve asked? – GalAbra Feb 22 '18 at 22:42
  • Yeah, they are. – JonSnow Feb 22 '18 at 22:43

1 Answers1

0

ok, I was able to solve it now.

$("#btnBest").click(function () {

   var Price = [];
   var Duration = [];
   var Stops = [];

   divList.each(function() {

       Price.push($(this).data('price'));
       Duration.push($(this).data('duration'));
       Stops.push($(this).data('stops'));

   });


   for(i=0; i< divList.length; i++) {

       var PriceSum = 0;
       var DurationSum = 0;
       var StopsSum = 0;
       var OverallSum = 0;

       for(j=0; j < divList.length; j++) {

           var newPriceSum = Price[i] / Price[j];
           var newDurationSum = Duration[i] / Duration[j];
           var newStopsSum = Stops[i] / Stops[j];

           PriceSum = PriceSum + newPriceSum;
           DurationSum = DurationSum + newDurationSum;
           StopsSum = StopsSum + newStopsSum;
       }

       var OverallSum = PriceSum + DurationSum + StopsSum;
       $(".listing-item" ).eq( i ).attr("data-sum", OverallSum);

       //console.log(i + ": " + PriceSum + " + " + DurationSum + " + " + StopsSum + " = " + OverallSum);

    } 

    divList.sort(function(a, b) { 
        return $(a).data("sum")-$(b).data("sum")
    });
    $("#list").append(divList);

});

See working Fiddle, if interested. https://jsfiddle.net/SchweizerSchoggi/k3af7pb0/

JonSnow
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