Need a display image script that works for this insert image script

Well the subject line is pretty explicit. I found this script that uploads a picture onto a folder on the server called images, then inserts the the path of the image on the images folder onto a VACHAR field in a database table.
[php]

<?php //This file inserts the main image into the images table. //address error handling ini_set ('display_errors', 1); error_reporting (E_ALL & ~E_NOTICE); //authenticate user //Start session session_start(); //Connect to database require ('config.php'); //Check whether the session variable id is present or not. If not, deny access. if(!isset($_SESSION['id']) || (trim($_SESSION['id']) == '')) { header("location: access_denied.php"); exit(); } else{ // Check to see if the type of file uploaded is a valid image type function is_valid_type($file) { // This is an array that holds all the valid image MIME types $valid_types = array("image/jpg", "image/jpeg", "image/bmp", "image/gif"); if (in_array($file['type'], $valid_types)) return 1; return 0; } // Just a short function that prints out the contents of an array in a manner that's easy to read // I used this function during debugging but it serves no purpose at run time for this example function showContents($array) { echo "
";

    		print_r($array);

   		 echo "
"; } // Set some constants // This variable is the path to the image folder where all the images are going to be stored // Note that there is a trailing forward slash $TARGET_PATH = "images/"; // Get our POSTed variable $image = $_FILES['image']; // Sanitize our input $image['name'] = mysql_real_escape_string($image['name']); // Build our target path full string. This is where the file will be moved to // i.e. images/picture.jpg $TARGET_PATH .= $image['name']; // Make sure all the fields from the form have inputs if ( $image['name'] == "" ) { $_SESSION['error'] = "All fields are required"; header("Location: member.php"); exit; } // Check to make sure that our file is actually an image // You check the file type instead of the extension because the extension can easily be faked if (!is_valid_type($image)) { $_SESSION['error'] = "You must upload a jpeg, gif, or bmp"; header("Location: member.php"); exit; } // Here we check to see if a file with that name already exists // You could get past filename problems by appending a timestamp to the filename and then continuing if (file_exists($TARGET_PATH)) { $_SESSION['error'] = "A file with that name already exists"; header("Location: member.php"); exit; } // Lets attempt to move the file from its temporary directory to its new home if (move_uploaded_file($image['tmp_name'], $TARGET_PATH)) { // NOTE: This is where a lot of people make mistakes. // We are *not* putting the image into the database; we are putting a reference to the file's location on the server $sql = "insert into images (member_id, image_cartegory, image_date, image) values ('{$_SESSION['id']}', 'main', NOW(), '" . $image['name'] . "')"; $result = mysql_query($sql) or die ("Could not insert data into DB: " . mysql_error()); header("Location: images.php"); echo "File uploaded"; exit; } else { // A common cause of file moving failures is because of bad permissions on the directory attempting to be written to // Make sure you chmod the directory to be writeable $_SESSION['error'] = "Could not upload file. Check read/write persmissions on the directory"; header("Location: member.php"); exit; } } //End of if session variable id is not present. ?>

[/php]
The script seems to work fine because I managed to upload a picture which was successfully inserted into my images folder and into the database.
Now the problem is, I can’t figure out exactly how to write the script that displays the image on an html page. I used the following script which didn’t work.

[php]
//authenticate user
//Start session
session_start();

    //Connect to database

require (‘config.php’);

$sql = mysql_query("SELECT* FROM images WHERE member_id = '".$_SESSION['id']."' AND image_cartegory = 'main' ");
$row = mysql_fetch_assoc($sql);
$imagebytes = $row['image'];
header("Content-type: image/jpeg");
print $imagebytes;

[/php]
Seems to me like I need to alter some variables to match the variables used in the insert script, just can’t figure out which. Can anyone help??

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