Load .php page into a div Ajax and Javascript

Question

(I am pretty new to all this.) I have a page where the user can click between different sorting methods of user posts. Depending on which they click the content shown will correspond.

function newContent()
{
  $.ajax({
      url: 'ajax/newcontent.php',
      success: function() 
      {
            $("#content_div").empty().load("ajax/newcontent.php");
      }

      });
}

This is what I have now. But instead of loading the code from the .php page into the #content_box, the #content_box just goes blank.

Any help would be awesome!

EDIT ** changed #content_dov to #content_div - whoops

you should give parameter to success function like success: function(msg) { $("#content_dov").empty().html(msg); } — Ricky
– Ricky, Commented Jul 16, 2012 at 4:58
.load() is shorthand for an entirely new ajax request. It automatically overwrites the contents of it's target, so you don't need to call empty. — nbrooks
– nbrooks, Commented Jul 16, 2012 at 5:03

Joyce Babu · Accepted Answer · 2012-07-16 04:59:17Z

5

Why are you using two ajax requests? You just need to use

$("#content_dov").load("ajax/newcontent.php");

The full code is

function newContent()
{
    $("#content_dov").load("ajax/newcontent.php");
}

answered Jul 16, 2012 at 4:59

Joyce Babu

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Comments

Rab Khan · Accepted Answer · 2012-07-16 04:58:50Z

3

function newContent()
{
  $.ajax({
      url: 'ajax/newcontent.php',
      success: function(data) 
      {
            $("#content_dov").html(data);
      }

      });
}

answered Jul 16, 2012 at 4:58

Rab Khan

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1 Comment

dabious Over a year ago

Both answers worked, but @Joyce Babu answer came first. Thanks!

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