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TiiTii

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  1. 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> <?php define('DB_NAME','xxxxxx'); define('DB_USER','xxxxxxxxx'); define('DB_PASSWORD','xxxxxxxx'); define('DB_HOST','xxxxxxxx.com'); //Ansluta mysql_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: <?php define('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!
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