Question

Myproblem is when I click my first selectbox, after there is a value, it will trigger the second selectbox. I implement it in Ajax, but successfully render ,but my other textfield value is gone. How could I just render a specific part of the responde html(success ajax call)?

$(document).ready(function(){ 

    if ($('#product_category').val() == 'Choose Category')
        document.getElementById('product_subcategory').disabled = true;

    $('#product_category').change(function () {   
        if ($('#product_category').val() == 'Choose Category')
            document.getElementById('product_subcategory').disabled = true;
        else
            document.getElementById('product_subcategory').disabled = false;


        data = $('#product_category').val();
        //alert(data);

        var param = 'category_name=' + data;
        $.ajax({
          url: MYURL,
          data: param,
          success: function(result) {
            alert('Choose product subcategory');
            alert(param);
            $('body').html('');
            $('body').html(result);
          }
        });
       //   window.location = MYURL?category_name="+data;
    });

    $('#product_subcategory').change(function () {    
        data = $('#product_subcategory').val();
     //     paramCategory = $(document).getUrlParam('category_name');
      //    alert(paramCategory);

        $.get(MYURL, function(data){
            alert("Data Loaded: " + data);
            });
        //window.location = MYURL?subcategory_name=" +  data;
    });

});

in my form, i use $_GET[‘category_name’] to get my value Ajax return value. I Debug in firebug, and it is successfully. I tried to render again the html, but my previous textarea’s value and textfiel’s value is gone since what i did is $(‘body’).html(”); $(‘body’).html(result);, So,how could I manage to get the success ajax return value, and use it in the PHP.

any confusion ,please tell me…
Thank you for spending ur time.

---

Hmm, I’m using a div and show the Div when it was return success ajax call.

Answer 1

You have two ways to do it.

1. Add the text from PhP while returning the data for Ajax call

2. Add the data back to the text box after loading the page..

   data = $(‘#product_category’).val();
   //alert(data);

   var param = ‘category_name=’ + data;
   $.ajax({
   url: MYURL,
   data: param,
   success: function(result) {
   alert(‘Choose product subcategory’);
   alert(param);
   $(‘body’).html(”);
   $(‘body’).html(result);
   $(‘#product_category’).val(data);
   }
   });