jquery/php form in modal window - php

Question

I have a form in a modal window. When I submit the form through ajax I don't get the success message. My aim is to see the message created in the php file in the modal after submitting the form. Here is the code:
<p><a class='activate_modal' name='modal_window' href='#'>Sign Up</a></p>
<div id='mask' class='close_modal'></div>
<div id='modal_window' class='modal_window'>
<form name="field" method="post" id="form">
<label for="username">Username:</label><br>
<input name="username" id="username" type="text"/><span id="gif"><span>
<span id="user_error"></span><br><br>
<label for="email">Email:</label><br>
<input name="email" id="email" type="text"/><span id="gif3"></span>
<span id="email_error"></span><br><br>
<input name="submit" type="submit" value="Register" id="submit"/>
</form>
</div>
The modal.js
$('.activate_modal').click(function(){
var modal_id = $(this).attr('name');
show_modal(modal_id);
});
$('.close_modal').click(function(){
close_modal();
});
$(document).keydown(function(e){
if (e.keyCode == 27){
close_modal();
}
});
function close_modal(){
$('#mask').fadeOut(500);
$('.modal_window').fadeOut(500);
}
function show_modal(modal_id){
$('#mask').css({ 'display' : 'block', opacity : 0});
$('#mask').fadeTo(500,0.7);
$('#'+modal_id).fadeIn(500);
}
The test.js for the registration of the user
$(function() {
$('#form').submit(function() {
$.ajax({
type: "POST",
url: "test.php",
data: $("#form").serialize(),
success: function(data) {
$('#form').replaceWith(data);
}
});
});
});
And the PHP FILE
<?php
$mysqli = new mysqli('127.0.0.1', 'root', '', 'project');
$username = $_POST['username'];
$email = $_POST['email'];
$mysqli->query("INSERT INTO `project`.`registration` (`username`,`email`) VALUES ('$username','$email')");
$result = $mysqli->affected_rows;
if($result > 0) {
echo 'Welcome';
} else {
echo 'ERROR!';
}
?>

Answer 1

Try putting the returncode from your AJAX call into
$('#modal_window')
instead of in the form
$('#form')
BTW: Why not use the POST or GET method of jQuery? They're incredibly easy to use...

Answer 2

Try something like this.
First write ajax code using jquery.
<script type="text/javascript">
function submitForm()
{
var str = jQuery( "form" ).serialize();
jQuery.ajax({
type: "POST",
url: '<?php echo BaseUrl()."myurl/"; ?>',
data: str,
format: "json",
success: function(data) {
var obj = JSON.parse(data);
if( obj[0] === 'error')
{
jQuery("#error").html(obj[1]);
}else{
jQuery("#success").html(obj[1]);
setTimeout(function () {
jQuery.fancybox.close();
}, 2500);
}
}
});
}
</script>
while in php write code for error and success messages like this :
if(//condition true){
echo json_encode(array("success"," successfully Done.."));
}else{
echo json_encode(array("error","Some error.."));
}
Hopes this help you.