jquery - Grouping by matching div value with query -

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i getting total of div values fine, not sure how breakdown of values. how many of each div value match. static example under breakdown. 

<h2>prices</h2> <div id="price1" class="price">125.95</div> <div id="price2" class="price">312.00</div> <div id="price3" class="price">560.00</div> <div id="price4" class="price">100.00</div> <div id="price5" class="price">125.95</div> <div id="price6" class="price">100.00</div> <div id="price7" class="price">125.95</div> <div id="price8" class="price">560.00</div> <div id="price9" class="price">100.00</div> <div id="price10" class="price">100.00</div> <div id="totalprice"></div> <h2>breakdown</h2> <div id="breakdown"> <p>3 prices @ 100.00</p> <p>3 prices @ 125.95</p> <p>1 price @ 312.00</p> <p>2 prices @ 560.00</p> </div>      <script type="text/javascript">              $(document).ready(function(e) {  //sum prices                     var sum = 0;                     $('.price').each(function(){                          sum += parsefloat($(this).text());                       });                          console.log('total: ' + sum.tofixed(2));   // breakdown                     $('.price').each(function(){                         sum = parsefloat($(this).text());                           console.log(sum.tofixed(2));                     });              });     </script>

similar above answer, did fiddle ¯\_(ツ)_/¯

var mappedvalues = {}; var breakdownsel = $('#breakdowns');  $('.price').each(function(index, value) {     var itemvalue = $(value).text();     mappedvalues[itemvalue] ? mappedvalues[itemvalue]++ : mappedvalues[itemvalue] = 1; });  (var prop in mappedvalues) {     if (mappedvalues.hasownproperty(prop)) {       breakdownsel.append('<p>' + mappedvalues[prop] + ' prices @ ' + parsefloat(prop).tofixed(2) + '</p>');     } };

see live: https://jsfiddle.net/v7vrz18p/

a few optimizations, namely check hasownproperty object's prototype contains additional properties technically still part of object, not necessary loop through in case.


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