help with mysql result resource error!

Anon

why am i getting this error from my code?

Warning: Supplied argument is not a valid MySQL result resource in c:\Program Files\Apache Group\Apache\htdocs\tastywet\order_func.php on line 32

from the line of code:

$conn = db_connect();

//insert clients address
$query = "select cid from clients where
cmpnyname = '$cmpnyname' fname = '$fname'
and lname = '$lname' and house_no = '$house_no'
and street = '$street' and town = '$town'
and postcode = '$postcode' and email = '$email'
and contact_no = '$contact_no'";
$result = mysql_query($query);
if(mysql_numrows($result)>0)//error is said to be here?
{
$cid = mysql_result($result, 0, "cid");
}
else
{
$query = "insert into clients values
('', '$cmpnyname', '$fname', '$lname', '$house_no',
'$street','$town','$postcode','$email','$contact_no')";
$result = mysql_query($query);
if (!$result)
return false;
}

Anon

Try this:

$query = "select cid from clients where ";
$query .= "cmpnyname = '$cmpnyname' ";
$query .= "AND fname = '$fname' ";
$query .= "and lname = '$lname' ";
$query .= "and house_no = '$house_no' ";
$query .= "and street = '$street' ";
$query .= "and town = '$town' ";
$query .= "and postcode = '$postcode' ";
$query .= "and email = '$email' ";
$query .= "and contact_no = '$contact_no'";

Also, are you sure house_no should be in quotes? From the name, I take it is a number...Same thing for contact_no, and maybe postcode.

Diego

Anon

yeh but i have them as varchars as house numbers can also have names instead of numbers! yeh? anyway I need to select all of the information from the database to see if that client already exists, if not then insert all of the details! How do I do this from what i gave you before?

ps whats the difference between:

"insert into clients values ('".$fname."')";

and

"insert into clients values ('$fname')";

my examples do not say why they can be different!

thanks!

Anon

The code you posted seems to do what you describe...

Regarding your P.S., there's no difference.

Diego