ajax image upload and preview form - php

Question

Im using a script to out put uploaded image preview. It works fine. i just like to show the image in one div and error or success massage in another div. is there any chance to do this? Here is the code.
Java script
<script type="text/javascript">
$(document).ready(function(){
$('#photoimg').live('change', function(){
$("#preview").html('');
$("#preview").html('<img src="loader.gif" alt="Uploading...."/>');
$("#imageform").ajaxForm(
{
target: '#preview'
}).submit();
});
});
</script>
HTML Code
<?php
include('db.php');
session_start();
$session_id='1'; // User login session value
?>
<div id="output"></div>
<form id="imageform" method="post" enctype="multipart/form-data" action='ajaximage.php'>
Upload image <input type="file" name="photoimg" id="photoimg" />
</form>
<div id='preview'>
</div>
PHP Code
include('db.php');
session_start();
$session_id='1'; // User session id
$path = "uploads/";
$valid_formats = array("jpg", "png", "gif", "bmp","jpeg");
if(isset($_POST) and $_SERVER['REQUEST_METHOD'] == "POST")
{
$name = $_FILES['photoimg']['name'];
$size = $_FILES['photoimg']['size'];
if(strlen($name))
{
list($txt, $ext) = explode(".", $name);
if(in_array($ext,$valid_formats))
{
if($size<(1024*1024)) // Image size max 1 MB
{
$actual_image_name = time().$session_id.".".$ext;
$tmp = $_FILES['photoimg']['tmp_name'];
if(move_uploaded_file($tmp, $path.$actual_image_name))
{
mysql_query("UPDATE users SET profile_image='$actual_image_name' WHERE uid='$session_id'");
echo "<img src='uploads/".$actual_image_name."' class='preview'>";
echo "<span class=ok-msg">Image has been uploaded..!</span>";
}
else
echo "<span class=error-msg">failed<span>";
}
else
echo "<span class=error-msg">Image file size max 1 MB</span>";
}
else
echo "<span class=error-msg">Invalid file format..</span>";
}
else
echo "<span class=error-msg">Please select image..!</span>";
exit;
}
I like to show all the massages (error-msg, ok-msg) and in the div output and image in the same place div preview. can anyone tell me how to do this. thanks in advance.

Answer 1

Use json dataType and success callback function like,
$("#imageform").ajaxForm(
{
dataType:'json',
success:function(json){
$('#output').html(json.img);
$('#preview').html(json.msg);
}
}).submit();
In PHP
Return data like
echo json_encode(array('img'=>"<img src='...' />",'msg'=>"Message goes here"));
return;

Answer 2

you can use the -success: and the error: eventListner
$("#imageform").ajaxForm(
{
dataType:'json',
success:function(json){
write the success msg
}
error: function() {
write the error msg
}
}).submit();