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Thanks for the response, DavidM and lakshmiyb I tried stripslashes() with existing and new entires into the database and mysql_real_escape() with new ones but neither worked I'll keep trying to get it working, thanks again for the help!

Ok, so first off would like to say Hello So I've been trying to figure out how to store images in a mysql database, and as far as i can tell the images are stored but getting them out seems to be the problem. when i try to go to the page on my webhost it says that "Can not select the database: Access denied for user 'testimg'@'localhost' to database 'testImages'" and when i goto show.php?id=1 on my local mamp install it gives me all kinds of weird symbols ÿÀ�™Ì�ÿÄ�È���������������������������!1AQaq"2‘¡±BR#Ábr‚’Ñ¢Â3CS$á²Òsƒ“%ðñc£Ã4â³T56Dt”E&Ód¤´ÄUu•7���!1AQaqð‘¡"2±ÁÑáñBRÂbr‚¢â#3ÿÚ���?�Ü/é -è i've checked everything in the code a million times and searched google and this forum for anything that can help me but i haven't been able to find something that has helped me understand what exactly is going on and why heres the upload form <form enctype="multipart/form-data" action="insert.php" method="post" name="changer"> <input name="MAX_FILE_SIZE" value="1500000" type="hidden"> <input name="image" accept="image/jpeg" type="file"> <input value="Submit" type="submit"> </form> this is my insert.php that processes the image after its submitted <?php include './database.php'; $link = mysql_connect($host, $username, $password); if (!$link) { die('Could not connect: ' . mysql_error()); } mysql_select_db ($database); if (isset($_FILES['image']) && $_FILES['image']['size'] > 0) { $tmpName = $_FILES['image']['tmp_name']; $fp = fopen($tmpName, 'r'); $data = fread($fp, filesize($tmpName)); $data = addslashes($data); fclose($fp); $query = "INSERT INTO tbl_images "; $query .= "(image) VALUES ('$data')"; $results = mysql_query($query, $link); print "Thank you, your file has been uploaded."; }else{ print "No image selected/uploaded"; } mysql_close($link); ?> and heres the show.php that displays the image using the id <?php include './database.php'; @mysql_connect($host, $username, $password) or die("Can not connect to database: ".mysql_error()); @mysql_select_db($database) or die("Can not select the database: ".mysql_error()); $id = $_GET['id']; if(!isset($id) || empty($id)){ die("Please select your image!"); }else{ $query = mysql_query("SELECT image FROM tbl_images WHERE id='".$id."'"); $row = mysql_fetch_assoc($query); $content = $row['image']; header("Content-type: image/jpg"); echo $content; } ?>