Jquery ajax php only working once

Question

I have the following jQuery

        $("button#submit").click(function(){
            $.ajax({
                type: "POST",
                url: "process.php",
                data: $('form.contact').serialize(),
              success: function(msg){
                $("#form-content").modal('hide');                     
                $("#thanks").html(msg);
                $("#thanks").delay(2000).fadeOut("slow");
              },
                error: function(){
                    alert("failure");
                }
            });
        });

And Php

<?php
    if (isset($_POST['name'])) {
    $name = strip_tags($_POST['name']);
    $email = strip_tags($_POST['Email']);
    $sug = strip_tags($_POST['sug']);

    echo "<span class='label label-info'>Your Website has been submitted .. Thank you</span>";
}?>

This works the first time and displays the php echo on my page. But when I submit the form again it does not show.

Your $("#thanks") is hidden

viclim
– viclim

2013-04-01 04:08:48 +00:00
Commented Apr 1, 2013 at 4:08 — viclim
– viclim, Commented Apr 1, 2013 at 4:08
its better if u can put this in a jsfiddle

Swarne27
– Swarne27

2013-04-01 04:09:14 +00:00
Commented Apr 1, 2013 at 4:09 — Swarne27
– Swarne27, Commented Apr 1, 2013 at 4:09

Dipesh Parmar · Accepted Answer · 2013-04-01 05:13:59Z

5

Your $("#thanks") dom is hidden.

    $("button#submit").click(function(){
        $.ajax({
            type: "POST",
            url: "process.php",
            data: $('form.contact').serialize(),
          success: function(msg){
            $("#form-content").modal('hide');                     
            $("#thanks").html(msg);
            $("#thanks").show();  <----------------ADD THIS
            $("#thanks").delay(2000).fadeOut("slow");
          },
            error: function(){
                alert("failure");
            }
        });
    });

edited Apr 1, 2013 at 5:13

Dipesh Parmar

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answered Apr 1, 2013 at 4:10

viclim

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Comments

Dipesh Parmar · Accepted Answer · 2013-04-01 05:47:55Z

2

Use .on().

$(document).on('click','button#submit',function(){  })

edited Apr 1, 2013 at 5:47

answered Apr 1, 2013 at 4:10

Dipesh Parmar

27.3k8 gold badges65 silver badges92 bronze badges

6 Comments

Ram Over a year ago

There is no difference between the on('click') and click.

Dipesh Parmar Over a year ago

@undefined i am afraid bro but there is....on('click') word as delegate and work on dynamically loaded content.

Ram Over a year ago

No, this not how on is used for event delegation, it's just a regular handler.

Dipesh Parmar Over a year ago

@undefined as far as i know its behave like i said....if i am wrong please correct me so i will not make mistake in future

Ram Over a year ago

$('#staticElement').on('click', '.dynamicallyGeneratedElement', function(){})

|

Community · Accepted Answer · 2017-05-23 12:21:52Z

-1

use like this

$("button#submit").live('click', function(e) {
  $.ajax({
            type: "POST",
            url: "process.php",
            data: $('form.contact').serialize(),
          success: function(msg){
            $("#form-content").modal('hide');                     
            $("#thanks").html(msg);
            $("#thanks").delay(2000).fadeOut("slow");
          },
            error: function(){
                alert("failure");
            }
        });
    });

check difference here .live and .on jQuery .live() vs .on() method for adding a click event after loading dynamic html

edited May 23, 2017 at 12:21

CommunityBot

11 silver badge

answered Apr 1, 2013 at 4:12

M.I.T.

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