CODEWITHSUNDEEP
jqueryajax
lgt
lgt

Reputation: 1034

how to render jquery ajax response

I'm quite new to jquery ajax and I'm trying to figure out how can I render the response data what I get back from php as json to update a specified div. So practically I'm having the following issue.

The JavaScript:

<script type="text/javascript">
  $(document).ready(function() {
    $(".ajax_call").change(function() {
      var domain = document.domain;
      var count = $('.ajax_call :selected').val();
      var $parent = $(this).closest(".product_box");
      var modul_title = $("h4", $parent).text();
      $.ajax({
        url:'index/ajax',
        data:{mod_title:modul_title, domain:domain, count:count},
        cache:'false',
        datatype:'json',
        success: function(response) {
          if (response.status = modul_title) {
            $parent.fadeOut();
            $parents.(response).fadeIn();                            
          } else {
            alert("Oops, script is a no go");
          }    
        }   
      });
   });
});

And the HTML:

<div class="product_box">
  <h4><!-- php code generating --> header</h1>
  <div class="product">
    <div class="thumbnail-item">
      <a href=""></a>
      <!-- and couple of other divs what are rendering my output in my mvc view -->
      <div class="ajax_bar">
        <!-- and here comes the dropdown what is triggering an ajax call -->
        <select class="ajax_call" size="1" name="blala">
          <option value='50'>add more 50</option>
          <option value='100'>add more 100</option>
          <option value='150'>add more 150</option>
        </select>
      </div>
    </div>
  </div>
</div>

What I want is to update the number of presented items with an ajax call. What is not clear to me how to render the response. Do I have to form the raw data again I mean as my html-php code looks or it can be done in another way?

Upvotes: 1

Views: 3510

Answers (3)

dhinesh
dhinesh

Reputation: 4764

Suppose you want to generate userdefined html then can use microtemplating MicroTemplating

Hope this is what you want.

Upvotes: 1

Pranay Rana
Pranay Rana

Reputation: 176906

you need to user html or text method for this to dispaly

$parents.html(response).fadeIn();

or 

$parents.text(response).fadeIn();

instead of $parents.(response).fadeIn();

Upvotes: 0

thecodeparadox
thecodeparadox

Reputation: 87073

$parents.(response).fadeIn();

should be

$parents.html(response).fadeIn();

or 

$parents.text(response).fadeIn();

.html() update the content of $parents

read about .text()

Upvotes: 0

Related Questions