Openlayers 3，从所选功能获取id

Question

我有这个图层

       $.getJSON("/g_estados/GetPrediosJson", null, function (data)
       {

           var format = new ol.format.WKT();

           var feature = new Array();

           for (i = 0, len = data.length; i < len; i++)
           {
               var fea = format.readFeature(data[i].wkt
                   , {
                       dataProjection: 'EPSG:3857',
                       featureProjection: 'EPSG:3857',
                       id: data[i].clave_catastral_g
                   }
           );
               //fea.attributes
               if (i < 10) {
                   console.log(fea)
               }
               feature.push(fea);
           }


           c_estados = new ol.layer.Vector({
               name: 'estados',
               source: new ol.source.Vector({
                   features: feature
               }),
               style: new ol.style.Style({
                   fill: new ol.style.Fill({
                       color: 'rgba(255, 255, 255, 0.2)'
                   }),
                   stroke: new ol.style.Stroke({
                       color: '#ffcc33',
                       width: 2
                   }),
                   image: new ol.style.Circle({
                       radius: 7,
                       fill: new ol.style.Fill({
                           color: '#ffcc33'
                       })
                   })

               })
           });

和一些选择这样的选项：

  

featuresSelect = new ol.Collection（）;

       select = null;  // ref to currently selected interaction
       // select interaction working on "singleclick"
       var selectSingleClick = new ol.interaction.Select({ features: >featuresSelect });

      // select interaction working on "click"
       var selectClick = new ol.interaction.Select({
           condition: ol.events.condition.click,
           features: featuresSelect
       });

       // select interaction working on "pointermove"
       var selectPointerMove = new ol.interaction.Select({
           condition: ol.events.condition.pointerMove,
           features: featuresSelect
       });

       var selectAltClick = new ol.interaction.Select({
           condition: function (mapBrowserEvent)
           {
               return ol.events.condition.click(mapBrowserEvent) &&
                   ol.events.condition.altKeyOnly(mapBrowserEvent);
           }
       });

       var changeInteraction = function ()
       {
           if (select !== null)
           {
               map.removeInteraction(select);
           }
           var value = 'singleclick';
           if (value == 'singleclick')
           {
               select = selectSingleClick;
           } else if (value == 'click')
           {
               select = selectClick;
           } else
           {
               select = null;
           }
           if (select !== null)
           {
               map.addInteraction(select);
               select.on('select', function (e)
               {

                   document.getElementById('status').innerHTML = ' ' >
                       e.target.getFeatures().getLength() +
                       ' selected features (last operation selected ' + >e.selected.length +
                       ' and deselected ' + e.deselected.length + ' >features)';
                   //document.getElementById('id01').style.display = >'block';
                   //console.log(e.selected);
                   //console.log(e.target.getFeatures()[0]);
                   console.log(featuresSelect);
                   //console.log(featuresSelect[0]);


                   var features2 = new Array;
                   feature2 = featuresSelect;

                   console.log(features2)

                   for (i = 0, len = featuresSelect.getLength() ; i < len; >i++)
                   {
                       console.log(featuresSelect[i]);
                   }
                   //{

所以在我选择一个功能之后我想知道基于这个id调用其他函数的id，但是我已经被困在这里几个小时了，openlayers的样本和文档都没有得到我的支持。

注意：我不寻求jquery（我不知道）或其他框架，只是纯粹的javascript，html，css。

我正在使用最新版本的openlayers，感谢您的帮助。