Crazy result resource error with mysql

[slighltywhacked]

Hey, new to the php/mysql scene and if possible could I get a little help?

Im gettin this error

Warning: Supplied argument is not a valid MySQL result resource in /www/welcometovernon/coupons/update.php on line 9

line 9 reads...
$num = mysql_num_rows($result);


everything up to that point..

@mysql_connect("localhost", "welcometovernon", "bob123")
or die("Unable to connect");
$result = @mysql_db_query("coupons", "SELECT * FROM coupons WHERE username = '$username' AMD (password = '$password')");

$num = mysql_num_rows($result);

password and username are coming from a page before this in a form..

Any ideas?

thanx
talk to you later

[freakysid]

The error is caused because your call to mysql_db_query() failed to return a result set identifier.

Some problems:

1) int mysql_db_query (string database, string query [, int link_identifier])
From the signature you can see that the first arguement is the database name. Is coupons the name of the database or just a table. In most cases, a script only needs to access one database. Therefor, I usually see ppl write their code like this:

$conn = @mysql_connect("localhost", "welcometovernon", "xxxxxx") 
   or die("Unable to connect"); 
@mysql_select_db("dbname", $conn) 
   or die( "Unable to connect");

$result = mysql_query("SELECT * FROM coupons WHERE username = '$username'
                                         AND password = '$password'"); 
$num = mysql_num_rows($result); 

2) AMD make computer chips, you want AND

[slighltywhacked]

"coupons" is a database that has the table "coupons" in it

[freakysid]

Well most likely there is a problem with your SQL statement. Insert this line of code somewhere:

echo "SELECT***FROM*coupons*WHERE*username*=*'$username'   
   AND*password*=*'$password'"; 

And see if it prints the SQL statement you expect to your browser. Also, you can then copy and paste this from your browser and run the sql from the command line (or from phpMyAdmin) and see what the result is.