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