concatenating two variable into one

Question

I'm new here and kind pretty new to PHP and I'm no sure as to why this is not working.

If I echo $ordernum1, I get the value I look for, but echoing echo ${"ordernum".$x}; gives me nothing.

I have also echoed $attempts and I get the value I'm looking for. Any help would be great thank you

$update=$_POST['update']; //echo $update;
$attempts=$_POST['attempts'];//echo $attempts;

if($update==2){
    for($x=0; $x<=$attempts; $x++){
        ${"ordernum".$x} = $POST["ordernum".$x.""]; echo ${"ordernum".$x};
        $query="UPDATE OrderTrack
        SET applicationID='--junk($appid)'
        WHERE OrderNum='".${"ordernum".$x}."'";
    }
}

Answer 1

Just $ordernum.$x is all that is required to append one var to another

Answer 2

You missed _ in

${"ordernum".$x} = $_POST["ordernum".$x.""];
             here --^

Answer 3

Try

$update = $_POST['update']; //echo $update; $attempts=$_POST['attempts'];//echo $attempts;

if ($update == 2) {
  for ($x = 0; $x <= $attempts; $x++) {
    $ordernum = $_POST["ordernum" . $x];
    echo $ordernum;
    $query = "UPDATE OrderTrack SET applicationID='--junk($appid)' WHERE OrderNum = '$ordernum'";
  }
}

You dont need to set a new variable in the loop, you can just reuse the $ordernum

Hope this helps

Luke

Answer 4

I wouldn't pollute your local scope with tons of related variables like this... Create an array to store that info

$ordernums = array();
if($update==2){
    for($x=0; $x<=$attempts; $x++){
        $ordernums[$x] = $_POST["ordernum".$x.""];

Not to mention that you have a massive SQL injection vulnerability in there. You need to escape any varaibles using something like mysql_real_escape_string() or mysqli::real_escape_string(). Or use a parameterized query (which is the best alternative).