How to debug using firebug and show the value of radio button

Question

I have radio buttons as shown below.

        


                        

        
                      

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Answer 1

In your javascript you are getting the value of the div, not the radio button:

$('#lensType').val() <--- change that

To something like this:

$("#lensType input:radio[name='design']:checked").val()

Answer 2

for debugging:

$('input[name="design"]').change(function(){ 
console.log($('#lensType').find("input:radio[name ='design']:checked").val());
});

else:

$('#lensType').find("input:radio[name ='design']:checked").val();

instead of

$('#lensType').val()

and you probably want to wrap it witg a changed function, since onload no design is selected:

  $(document).ready(function(){
$('input[name="design"]').change(function(){ 
var design = $('input[name="design"]:checked').val();
     function populate() {
      fetch.doPost('getSupplier.php');
   }

 $('#lensType').change(populate);

  var fetch = function() {

 var counties = $('#county');

return {
doPost: function(src) {

$('#loading_county_drop_down').show(); // Show the Loading...
$('#county_drop_down').hide(); // Hide the drop down
$('#no_county_drop_down').hide(); // Hide the "no counties" message (if it's the case)


    if (src) $.post(src, { supplier: design }, this.getSupplier);

    else throw new Error('No source was passed !');
},

getSupplier: function(results) {
    if (!results) return;
    counties.html(results);

$('#loading_county_drop_down').hide(); // Hide the Loading...
$('#county_drop_down').show(); // Show the drop down
}   
  }
 }();

 populate();
});
 });