Skip to main content
August 2, 2011
Question

jquery and php ajax

  • August 2, 2011
  • 1 reply
  • 651 views

Hi

I have a Front.php page which loads an external php page using:

<script type="text/javascript">
        $(document).ready(function() {

          //burnbackend is echoed to div commentfeed.
        $("#commentfeed").load("burnbackend.php");
               
            
            $("#shuffle").submit(function(){
                var voteone= $("#voteone").val();
                $.get("burnbackend.php", { voteone : voteone},  function(data) {
                    $("#commentfeed").html(data);
                    });      
                      
        });
</script>

//inside body

<div id="commentfeed"></div>


burnbackend.php is a page that echos a form the problem is that using jquery ajax I would like to submit the echoed form

to burnbackend.php  and load a different result.

burnbackend.php

<?Php

if($_GET['voteone'])

{

//Here the submit of the vote is done and some database stuff.

echo "vote done";

echo'

<form name="shuffle" id="shuffle" action="" method="get">
<input type="hidden" id="voteone" value="VOTE HERE"/><br/>
<input type="submit" id="submit" value="SHUFFLE BUTTON"/><br/>
</form>';

}

else

{

echo'

<form name="shuffle" id="shuffle" action="" method="get">
<input type="hidden" id="voteone" value="VOTE HERE"/><br/>
<input type="submit" id="submit" value="SHUFFLE BUTTON"/><br/>
</form>';

}

?>

The problem in a way goes like this: How do I submit an echoed form from a loaded external page using j query from the

page that loads the external page.

Thanks.

Please ask more questions if the problem is un understandable.

This topic has been closed for replies.

1 reply

pziecina
Legend
August 2, 2011

Hi

You include the link to the form processing script in the action element of the form -

<form name="shuffle" id="shuffle" action="" method="get">

I would also advise using post and not get as the method.

PZ

August 3, 2011

Hi thanks for reply

I changed assignments to post. I also linked the form to the Front.php that has the Jquery. Its posts

and a get a sucess but the current Front.php instantly changes to the backend.php which is

the page loaded into Front.php and contains the posting form: I need to do the post and keep Front.php but

just change the backend.php thats is load on Front.php

Code:

Front.php

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
<script type="text/javascript" src="jquery/jquery-1.4.2.min.js"></script>

<script type="text/javascript">
        $(document).ready(function() {
   
           
              $("#shuffle").submit(function() {
              var voteone= $("#voteone").val();
              $.post("burnbackend.php", { voteone : voteone},  function(data) {
                    $("#commentfeed").html(data);
              });
       
            });
           
           
           
        });
</script>
</head>

<body>

<div id="commentfeed"></div>
</body>
</html>

Backend.php


<?Php


$morecapture=$_GET['expid'];

$motion=$_POST['voteone'];

if($motion)
{
echo 'vote done';

echo'<form name="shuffle" id="shuffle" action="burnout.php" method="post">
<input type="hidden" id="voteone" value="VOTE HERE"/><br/>
<input type="submit" id="submit" value="SHUFFLE BUTTON"/><br/>
</form>';

}

?>

So as you can see. Front.php loads backend.php through Jquery. backend.php has an echoed form that I am trying to submit using

jquery from Front.php and changing what was original loaded from backend.php