Here is my problem. I am uploading images to my database but when I want to display them I cannot call the ID of the image. How do I get it to fetch the ID. If I specify the $id = 1 like so, it calls the image no problem when I use view.php?=1 etc but how do I get it so when is use the view.php?id=1 call it gets the id from the address? what have I done wrong or need to do? below is the code I have used to call the image thanks in advance
mysql_connect($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($id) || !is_int($id)){
die("Please select your image!");
}else{
$query = mysql_query("SELECT * FROM tbl_images WHERE id='".$id."'");
$row = mysql_fetch_array($query);
$content = $row['image'];
header('Content-type: image/jpg');
echo $content;
}
Don't use is_int()
, because it checks if the variable is of type integer, and $_GET
will always return a string, never an integer.
You want to use is_numeric
instead.