Variable math

[Modern Merlin]

I have been at this now for some time and while I know its probably s very simple answer I just cant seem to find it. What I am trying to do is take two numbers, which I convert to variables, then I subtract one from the other. Not sure which part is wrong but here is my code:

<?php
//--------------------MYSQL DETAILS ---------------------------------------//

//mysql details
require_once("config.php");

//Passing the lookup variable from flash
$user = $_POST['user'];
$stash = $_POST['tStash'];

$conn;
$SQL = "SELECT * FROM mobster WHERE account_id ='".$user."'";
$rs = mysql_query($SQL);

//Putting the results in an array
$result = mysql_fetch_array($rs);

//------------------ Assign Variables ---------------------------------//

$tTurns = $result['cturns'];
$tStashTurns = $result['stashTurns'];

if ($tStashTurns > 0){
$newStash = $tStashTurns - $stash;
$newTurns = $tTurns + $stash;
echo ("&result=success&");
} else {
echo "&result=failure&";
exit('&result=failure&'); //Abort PHP Script
}

$conn;
$insertSQL = "UPDATE mobster SET cturns='$newTurns', stashTurns='$newStash' WHERE account_id ='$user'";
$rs = mysql_query($insertSQL);

?>

Any ideas?

MM

[edited by: Modern_Merlin at 12:18 am (utc) on July 22, 2008]

[eelixduppy]

Code looks decent. Are you getting any errors? How about from MySQL? Try using mysql_error [www]().

[NomikOS]

you can do "something" like this:

if ($tStashTurns > 0) { 
  "UPDATE mobster SET cturns=cturns+$stash, stashTurns=stashTurns-$stash WHERE WHERE account_id ='$user'"; 
}

did you know it?