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Home/ Questions/Q 637881
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Editorial Team
  • 0
Editorial Team
Asked: May 13, 20262026-05-13T20:40:20+00:00 2026-05-13T20:40:20+00:00

I was given this code earlier but am having a hard time parsing the

  • 0

I was given this code earlier but am having a hard time parsing the correct data.

I have the following JSON

{
  flavors: [{
    "image": "images/bbtv.jpg",
    "desc": "BioBusiness.TV",
    "id": "1"
  }, {
    "image": "images/grow.jpg",
    "desc": "Grow Staffing",
    "id": "2"
  }]
}

and I want to only show id:1 or id:2.

I have the following code for Ajax

$.ajax({
        type: "POST",
        url: "foodservice/all.js",
        dataType: "json",
        cache: false,
        data: {"flavors": filterId(flavors, 'a')},
        contentType: "application/json",
        success: function(data) {
           $('#flavor-detail').html("<div/>");
           $.each(data.flavors, function(i,item){
              $('#flavor-detail div').append('<ul><li><p>' + item.desc + '</p></li></ul>');
           });
        }         
     });

and the following function to filter through the JSON object

function filterId(obj, filteredId) {
 var resultObj = $.extend({},obj);

 for (var i in obj) {
  if ( obj.hasOwnProperty(i) ) {
   if ( obj[i].id && obj[i].id !== filteredId ) {
    delete obj[i];
   }
  }
 }
 return resultObj;
}

However, this code does not return anything.
Can someone tell me what I am missing?

Im pretty new to JSON, Ajax so any help would be greatly appreciated.

Thanks!

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-13T20:40:20+00:00Added an answer on May 13, 2026 at 8:40 pm

    Why not just check in the “each” code?

      $.each(data.flavors, function(i,item){
          if (item.id > 2) return;
          $('#flavor-detail div').append('<ul><li><p>' + item.desc + '</p></li></ul>');
      });
    
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