I am creating a dynamic web page where uploading images is one of its feature. I have a looping DIV where each DIV has its own images: show sample below...
<?php
while($row=mysql_fetch_assoc($result)){
$id=$row['id'];
$path=$row['path'];
...
?>
<div class="img-container">
<img src="<?php echo $path; ?>" />
<button onclick="changeMe('<?php echo $id; ?>')">Change Photo</button>
</div>
...and i wanted the upload form be displayed in popup DIV. Is there anyone who could help me and give me the exact code for this because i am not so familiar on jquery and ajax. Where after processing the uploaded image the information will be sent back to the popup upload if it is successful or not. Any help for those who are knowledgable on this. Thanks!!