Jquery ajax request echo html data

Question

I'm making an ajax request to my PHP script which will echo out a certain # of tables ( different each time), that I want to include on my original page when a user clicks the submit_form button. My question is how can i get Jquery to display this table into the table_layout div?

HTML :

<div id="table_layout"> </div>

JQUERY

$( ".submit_form" ).click(function( e ) {


          $.ajax({
          type : 'GET',
          url:  'draw_tables.php',
          dataType : 'html',
          data:  dataString ,
          error : function(XMLHttpRequest, textStatus, errorThrown) {
          alert('error'); },
          success : function(data) { //appendto.('#table_layout')   } 
          });    
          return false; 
          })

PHP (draw_tables.php)

echo '<table>';
echo '<input type="text" name="ID">';
echo '<input type="text" name="CLASS">';
echo '</table>';

Between table_layout and what ?

Ricardo Lohmann
– Ricardo Lohmann

2012-12-06 20:39:10 +00:00
Commented Dec 6, 2012 at 20:39 — Ricardo Lohmann
– Ricardo Lohmann, Commented Dec 6, 2012 at 20:39
@RicardoLohmann sorry fixed left out the html part

Amanda Sky
– Amanda Sky

2012-12-06 20:40:47 +00:00
Commented Dec 6, 2012 at 20:40 — Amanda Sky
– Amanda Sky, Commented Dec 6, 2012 at 20:40
So, it's not between. It's in.

Ricardo Lohmann
– Ricardo Lohmann

2012-12-06 20:41:44 +00:00
Commented Dec 6, 2012 at 20:41 — Ricardo Lohmann
– Ricardo Lohmann, Commented Dec 6, 2012 at 20:41

rsz · Accepted Answer · 2012-12-06 20:48:13Z

2

For example you could make an ajax request and fill the div with it

 $.get("draw_tables.php", function(data) {
 $("#table_layout").html(data);
 });

This will fill your table_layout id with the data from the ajax request.

answered Dec 6, 2012 at 20:48

rsz

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Ricardo Lohmann · Accepted Answer · 2012-12-06 20:40:40Z

1

$('#table_layout').append(data);

answered Dec 6, 2012 at 20:40

Ricardo Lohmann

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1 Comment

Amanda Sky Over a year ago

the problem is im not returning any data back it's just echoing out the table from PHP

lortabac · Accepted Answer · 2012-12-06 20:57:53Z

0

Inside the success callback you should add:

$('#table_layout').append(data);

answered Dec 6, 2012 at 20:57

lortabac

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