views:

159

answers:

2

I am using php/ajax to submit a form without page refresh. Here are my files-

coupon.js

jQuery(document).ready(function(){
        jQuery(".appnitro").submit( function(e) {
$.ajax({
            url     : "http://174.132.194.155/~kunal17/devbuzzr/wp-content/themes/street/sms.php",
            type    : "post",
            dataType: "json",
            data    : $(this).serialize(),
            success : function( data ) {
                        for(var id in data) {
                            jQuery('#' + id).html( data[id] );
                        }
                      }

        });
//return false or
e.preventDefault();

    });

});

sms.php

    <?php
    //process form
$res = "Message successfully delivered";
    $arr = array( 'mess' => $res );
    echo json_encode( $arr );//end sms processing
    unset ($_POST);
    ?>

and here is code for my html page -

<form id="smsform" class="appnitro" action="http://174.132.194.155/~kunal17/devbuzzr/wp-content/themes/street/sms.php" method="post">
...
</form>
<div id="mess" style="background:green;"></div>

Now when i click on submit button nothing happens and firebug shows following under console panel -

POST http://174.132.194.155/~kunal17/devbuzzr/wp-content/themes/street/sms.php

404 Not Found 1.29s   `jquery.min.js (line 130)`

Response

Firebug needs to POST to the server to get this information for url:
http://174.132.194.155/~kunal17/devbuzzr/wp-content/themes/street/sms.php

This second POST can interfere with some sites. If you want to send the POST again, open a new tab in Firefox, use URL 'about:config', set boolean value 'extensions.firebug.allowDoublePost' to true
This value is reset every time you restart Firefox This problem will disappear when https://bugzilla.mozilla.org/show_bug.cgi?id=430155 is shipped

When i set 'extensions.firebug.allowDoublePost' to true then following results show up -

POST http://174.132.194.155/~kunal17/devbuzzr/wp-content/themes/street/sms.php

404 Not Found 1.29s   `jquery.min.js (line 130)`

Response - 

{"mess":"Message successfully delivered"}

CaN anyone help me in fixing this firebug error of 404 not found. And why is it showing jquery.min.js (line 130) along side?

P.S -do not worry about http://174.132.194.155/~kunal17/devbuzzr/wp-content/themes/street this is my base url

A: 

You may want to try putting the e.preventDefault() statement before the $.ajax call.

EDIT:

My x.html, corresponds to your HTML page


<!DOCTYPE html>
<html>
<head>
<title>x</title>
<script 
  type="text/javascript" 
  src="http://ajax.googleapis.com/ajax/libs/jquery/1.4.2/jquery.min.js"&gt;
</script>
<script type="text/javascript" src="/so/x.js"></script>
</head>
<body>
<form id="smsform" class="appnitro" action="/so/x.php">
<input type="text" name="zz">
<input type="submit">
</form>
<div id="mess" style="background:green;"></div>
</body>
</html>

My x.js, corresponds to your coupon.js


jQuery(document).ready(function(){
  jQuery(".appnitro").submit( function(e) {
    e.preventDefault();
    $.ajax({
      url     : "/so/x.php",
      type    : "post",
      dataType: "json",
      data    : $(this).serialize(),
      success : function( data ) {
        for(var id in data) {
          jQuery('#' + id).html( data[id] );
        }
      }

    });
    //return false or
    //e.preventDefault();
  });
});

My x.php, corresponds to your sms.php


<?php
$res = "Message successfully delivered.";
$arr = array('mess'=>$res);
echo json_encode($arr);
unset($_POST);
?>

This actually works in my environment, although I do not have the rest of the HTML markup or the additional PHP form processing code. The "Message successfully delivered." shows up in green directly below the input field.

jtp
@jt : no.does not help
ayush
Yep. Not root cause :) In my case it allowed the AJAX call to complete without the whole page having to refresh. Where is your jquery.min.js file being hosted?
jtp
A: 

When inside the Ajax call this refers to the Ajax object you need to do this

var __this = this; 

Before going into the Ajax call, then it would be

data    : __this.serialize()

Or look up the use of context within an Ajax call in Google. Or serialise your data into a variable before going into the Ajax call.

Liam Bailey