php error

[oakcha]

Am having an error on line 63 when a run this php code..
config.php

<?php
$mysql_hostname = "localhost";
$mysql_user = "root";
$mysql_password = "";
$mysql_database = "bankdb";
$bd = mysql_connect($mysql_hostname, $mysql_user, $mysql_password) 
or die("Opps some thing went wrong");
mysql_select_db($mysql_database, $bd) or die("Opps some thing went wrong");
?>


the error was found on this code

<?php require_once('config.php');?>
<?php
//initialize the session
if (!isset($_SESSION)) {
  session_start();
}

// ** Logout the current user. **
$logoutAction = $_SERVER['PHP_SELF']."?doLogout=true";
if ((isset($_SERVER['QUERY_STRING'])) && ($_SERVER['QUERY_STRING'] != "")){
  $logoutAction .="&". htmlentities($_SERVER['QUERY_STRING']);
}

if ((isset($_GET['doLogout'])) &&($_GET['doLogout']=="true")){
  //to fully log out a visitor we need to clear the session varialbles
  $_SESSION['MM_Username'] = NULL;
  $_SESSION['MM_UserGroup'] = NULL;
  $_SESSION['PrevUrl'] = NULL;
  unset($_SESSION['MM_Username']);
  unset($_SESSION['MM_UserGroup']);
  unset($_SESSION['PrevUrl']);
	
  $logoutGoTo = "mainpage.php";
  if ($logoutGoTo) {
    header("Location: $logoutGoTo");
    exit;
  }
}
?>
<?php
if (!function_exists("GetSQLValueString")) {
function GetSQLValueString($theValue, $theType, $theDefinedValue = "", $theNotDefinedValue = "") 
{
  if (PHP_VERSION < 6) {
    $theValue = get_magic_quotes_gpc() ? stripslashes($theValue) : $theValue;
  }

  $theValue = function_exists("mysql_real_escape_string") ? mysql_real_escape_string($theValue) : mysql_escape_string($theValue);

  switch ($theType) {
    case "text":
      $theValue = ($theValue != "") ? "'" . $theValue . "'" : "NULL";
      break;    
    case "long":
    case "int":
      $theValue = ($theValue != "") ? intval($theValue) : "NULL";
      break;
    case "double":
      $theValue = ($theValue != "") ? doubleval($theValue) : "NULL";
      break;
    case "date":
      $theValue = ($theValue != "") ? "'" . $theValue . "'" : "NULL";
      break;
    case "defined":
      $theValue = ($theValue != "") ? $theDefinedValue : $theNotDefinedValue;
      break;
  }
  return $theValue;
}
}
mysql_select_db($mysql_database, $bd);
$query_Recordset1 = "SELECT * FROM users";
$Recordset1 = mysql_query($query_Recordset1, $db) or die(mysql_error());
$row_Recordset1 = mysql_fetch_assoc($Recordset1);
$totalRows_Recordset1 = mysql_num_rows($Recordset1); 
?> 

from here down
mysql_select_db($mysql_database, $bd);
$query_Recordset1 = "SELECT * FROM users";
$Recordset1 = mysql_query($query_Recordset1, $db) or die(mysql_error());
$row_Recordset1 = mysql_fetch_assoc($Recordset1);
$totalRows_Recordset1 = mysql_num_rows($Recordset1); 
?>

[djr33]

1. What is the error (and can you point out the line, rather than just pasting the whole code)?
2. Post new questions in new threads.
3. Use [black] tags around code so it is easily readable.

[oakcha]

when a run with my browser i get this error
Warning: mysql_query(): supplied argument is not a valid MySQL-Link resource in C:\xampp\htdocs\alex\html\accountpage.php on line 63
i guess is from this part of the code

mysql_select_db($mysql_database, $bd);
$query_Recordset1 = "SELECT * FROM users";
$Recordset1 = mysql_query($query_Recordset1, $db) or die(mysql_error());.......................line 64
$row_Recordset1 = mysql_fetch_assoc($Recordset1);
$totalRows_Recordset1 = mysql_num_rows($Recordset1);

[Schmoopy]

The problem is that you've declared your database as $bd, not $db.

Change the mysql_query line to:

$Recordset1 = mysql_query($query_Recordset1, $bd) or die(mysql_error()); 

See if that fixes it