Php SQL syntax error

I've tri开发者_开发知识库ed to arrange this in a few ways but the error message stays almost the same:

15Error retrieving scores You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'table WHERE id>15 1 ORDER BY id ASC LIMIT 0,100' at line 1

The call i make is

http://myserver.com/get_dbupdates2.php?theid=15

    $type   = isset($_GET['type']) ? $_GET['type'] : "global";
$offset = isset($_GET['offset']) ? $_GET['offset'] : "0";
$count  = isset($_GET['count']) ? $_GET['count'] : "100";
$sort   = isset($_GET['sort']) ? $_GET['sort'] : "id ASC";

// Localize the GET variables
$udid  = isset($_GET['udid']) ? $_GET['udid'] : "";
$name  = isset($_GET['name']) ? $_GET['name']  : "";
$clubname  = isset($_GET['clubname']) ? $_GET['clubname']  : "";
$theid  = isset($_GET['theid']) ? $_GET['theid']  : ""; 


// Protect against sql injections
$type   = mysql_real_escape_string($type);
$offset = mysql_real_escape_string($offset);
$count  = mysql_real_escape_string($count);
$sort   = mysql_real_escape_string($sort);
$udid   = mysql_real_escape_string($udid);
$name   = mysql_real_escape_string($name);
$clubname   = mysql_real_escape_string($clubname);
$theid   = mysql_real_escape_string($theid);

    echo $theid;

// Build the sql query
//$sql = "SELECT * FROM $table WHERE ";
$sql = "SELECT * FROM $table WHERE id>$theid ";

switch($type) {
    case "global":
        $sql .= "1 ";
        break;
    case "device":
        $sql .= "udid = '$udid' ";
        break;
    case "name":
        $sql .= "name = '$name' ";
        break;
    case "clubname":
        $sql .= "clubname = '$clubname' ";
        break;
    case "theid":
        $sql .= "theid = '$theid' ";
        break;
}

$sql .= "ORDER BY $sort ";
$sql .= "LIMIT $offset,$count ";

$result = mysql_query($sql,$conn);

Anybody able to see where I went wrong?

Kindest Regards, -Code

EDIT

See these 2 lines

//$sql = "SELECT * FROM $table WHERE ";
$sql = "SELECT * FROM $table WHERE id>$theid ";

If i comment out the bottom line, and uncomment the top line the script runs ok and returns the data. But leaving it as it is gives the error.

So this leaves me to believe the problem is something to do with

id>$theid ";  

Regards -Code

Named table 'table' must be quoted like that

SELECT * FROM `table`

and you must define AND or OR between conditions e.g. $sql .= "AND clubname = '$clubname' ";