Hey there!
I have this big project due 5 days and I really need some help. I've made a imagesharing website where you upload images to a MySQL database into a table called images (with id, name and image)
This is the uploading form code:
<form action="laddaupp.html" method="POST" enctype="multipart/form-data"><p> Ladda upp här </p><input type="file" name="image" /> <input type="submit" value="Ladda upp" /></form><?phpdefine('DB_NAME','xxxxxx');define('DB_USER','xxxxxxxxx');define('DB_PASSWORD','xxxxxxxx');define('DB_HOST','xxxxxxxx.com');//Anslutamysql_connect(DB_HOST, DB_USER, DB_PASSWORD) or die (mysql_error());mysql_select_db(DB_NAME) or die (mysql_error());$file= $_FILES['image']['tmp_name'];if(!isset($file))echo "Välj en bild";else{$image = addslashes(file_get_contents($_FILES['image']['tmp_name']));$image_name = addslashes($_FILES['image']['name']);$image_size = addslashes(getimagesize($_FILES['image']['tmp_name']));if($image_size==FALSE)echo "Det är ingen bild";else{if(!$insert = mysql_query("INSERT INTO images VALUE ('','$image_name','$image')"))echo "Problem med uppladdning";else{$lastid = mysql_insert_id();echo "Klart! <p /> Bilden du laddade upp: <p /> <img src=get.php?id=$lastid>";}}}?>and this is the page which I'm trying to view the images on:<?phpdefine('DB_NAME','xxxxxxx');define('DB_USER','xxxxxxx');define('DB_PASSWORD','xxxxxx');define('DB_HOST','xxxxxxxxxxxxxx.com');mysql_connect(DB_HOST, DB_USER, DB_PASSWORD) or die (mysql_error());mysql_select_db(DB_NAME) or die (mysql_error());$res=mysql_query("SELECT * FROM images");echo "<table>";while($row=mysql_fetch_array($res)){echo "<tr>";echo "<td>";?> <img src="data:image/jpeg;base64,base64_encode($row['image'])"> <?php echo "</td>";echo "<td>"; echo $row["name"]; echo "</td>";echo "</tr>";}echo "</table>";mysql_close();?>When I open this file I get this:Why do my images get broken??Could someone please help me I'd be really happy!