Please help. This is what I have in my PHP code and it kepting giving me error.
<?PHP
//************ Variable Declaration ***************************************************
$sbSubmit = $HTTP_GET_VARS["btnSubmit"];
$txtUsrName = $HTTP_GET_VARS["txtUserName"];
$txtUsrPassword = $HTTP_GET_VARS["txtPassword"];
$self = $HTTP_SERVER_VARS['PHP_SELF'];
//************Testing and Find Records**************************************************
$db = mysql_connect("tsevneeglauj", "lorcz", "alcy123"
; //connect to database
if ($txtUsrName == "" || $txtUsrPassword == ""
{
echo "User name: ".$txtUsrName & " password: ".$txtUsrPassword;
echo "Please enter either your user name and/or password!";
exit;
}
elseif(!$db) {
echo "Error: Could not connect to database. Please try again later.";
exit;
}
mysql_select_db("losociety", $db);
$query = "SELECT * FROM users WHERE (usrName LIKE $txtUsrName AND usrPassword == $txtUsrPassword)";
$result = mysql_query($query, $db);
$num_results = mysql_num_rows($result);
echo "Num results: ".$num_results."<br>";
if ($num_results ==0){
echo"Either your password and/or user name is incorrect. Plesae try again.";
exit;
}
else{
require("resultTable.inc"
;
}
?>
The error is:
Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in D:\Inetpub\ on line 35
Many thanks in advance!
ljCharlie
<?PHP
//************ Variable Declaration ***************************************************
$sbSubmit = $HTTP_GET_VARS["btnSubmit"];
$txtUsrName = $HTTP_GET_VARS["txtUserName"];
$txtUsrPassword = $HTTP_GET_VARS["txtPassword"];
$self = $HTTP_SERVER_VARS['PHP_SELF'];
//************Testing and Find Records**************************************************
$db = mysql_connect("tsevneeglauj", "lorcz", "alcy123"

if ($txtUsrName == "" || $txtUsrPassword == ""

echo "User name: ".$txtUsrName & " password: ".$txtUsrPassword;
echo "Please enter either your user name and/or password!";
exit;
}
elseif(!$db) {
echo "Error: Could not connect to database. Please try again later.";
exit;
}
mysql_select_db("losociety", $db);
$query = "SELECT * FROM users WHERE (usrName LIKE $txtUsrName AND usrPassword == $txtUsrPassword)";
$result = mysql_query($query, $db);
$num_results = mysql_num_rows($result);
echo "Num results: ".$num_results."<br>";
if ($num_results ==0){
echo"Either your password and/or user name is incorrect. Plesae try again.";
exit;
}
else{
require("resultTable.inc"

}
?>
The error is:
Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in D:\Inetpub\ on line 35
Many thanks in advance!
ljCharlie