Hello there Good day to each and everyone of us here.
i have a problem displaying the image i have uploaded in the root directory folder. This is my codes in uploading image. I am using jquery.min.js and jquery.form.js.

$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
{
$actual_image_name = time().".".$ext;
$tmp = $_FILES['photoimg']['tmp_name'];
if(move_uploaded_file($tmp, $path.$actual_image_name))
{
mysql_query("UPDATE users SET profile_image='$actual_image_name' WHERE student_id='{$_SESSION['user_id']}'",$link_id);
echo "<img src='uploads/".$actual_image_name."'  class='preview'>";
}
else
echo "failed";
}
else
echo "Image file size max 1 MB"; 
}
else
echo "Invalid file format.."; 
}
else
echo "Please select image..!";
exit;
}
?>

i have tried using this script to display the image uploaded on the other page.

<div id="about-img">
<?php 
$actual_image_name = time().".";
mysql_query("UPDATE users SET profile_image='$actual_image_name' WHERE student_id='{$_SESSION['user_id']}'",$link_id);
echo "<img src='uploads/".$actual_image_name."'  class='profile-photo'>";
?>
</div>

but the image shows empty picture.
Please advise.

time() returns the actual time. So the time used in the upload, and the one for the display are different. You need the store the name you use for the upload, so you can use the exact same one when displaying.

So why couldn't you change the code here and continue this thread?