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Mysql error without a reason i can find

         

Twisted Mind

11:33 am on Oct 14, 2005 (gmt 0)

10+ Year Member



Here is my script and when i get the data from a database it gives this error:

Warning: mysql_fetch_row(): supplied argument is not a valid MySQL result resource in D:\Fluke_Mambo\components\com_confirmation\order_status.php on line 33

if you want to know where $my->username comes from it is defined from the user logged in at the moment becouse this file is included.

<?/************************************
**Order Confirmation form
**Created by MathijsSegers@gmail.com
**Copyright by FlukeCorp
************************************/

//Get Host
$Host ="localhost";

//Select Database
$Database ="fluke";

//Get database table
$table ="mos_pshop_orders";

//get 2cond table
$sessiontable="mos_session";

//get usertable
$usertable = "mos_users";

//Get database User
$User ="root";

//Get database Password
$Password ="#*$!xx";

$myusername = $my->username;
$connection = mysql_connect ($Host,$User,$Password);
$query = "SELECT departement, username, id FROM $usertable WHERE username == $my->username";
$result = mysql_db_query("$Database",$query);

while(list($departement, $username,$id)=mysql_fetch_row($result))
{
$username = $username;
$departement = $departement;
}
if ($my->username!= $username){
die;}


if ($departement == "it")
{
include"order_status_it.php";
}
if ($departement == "finance")
{
include"order_status_finance.php";
}
if ($departement == "manager")
{
include"order_status_manager.php";
}?>

plz omg plz help me

jatar_k

2:23 pm on Oct 14, 2005 (gmt 0)

WebmasterWorld Administrator 10+ Year Member



>> supplied argument is not a valid MySQL result resource

this means your query died, you need to properly trap the error from mysql on your query function

try this

$result = mysql_db_query("$Database",$query) or die (mysql_error());

that should echo the error from mysql to the browser

my bet your query is misconstructed

$query = "SELECT departement, username, id FROM $usertable WHERE username == $my->username";

it may just be the == in mysql you need to just use a sungle =

I am also not sure if you can use object notation inside double quotes and the username will need to be surrounded by single quotes. I usually concatenate my queries together like so

$query = "SELECT departement, username, id FROM " . $usertable . " WHERE username == '" . $my->username . "'";