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Warning: Wrong parameter count for mysql_result() in /home/h

Posted: Mon Oct 30, 2006 11:55 am
by oskare100
Hello,
When I rund this part of the script:

Code: Select all

$sql13 = "select account_username from paypal_sales where txn_id = '$txn_id'";
$result13 = mysql_query($sql13) or die( mysql_error() );
$del_account_username = mysql_result($result13);
echo "Username: $del_account_username";
I get the error:
Warning: Wrong parameter count for mysql_result() in /home/httpd/vhosts/com/

What does that mean? There is a value for account_username in the table paypal_sales where the txn_id in the database is the same as the $txn_id variable in the script.

Best Regards
Oskar R

Posted: Mon Oct 30, 2006 12:14 pm
by John Cartwright
The manual explains what parameters you need to pass

Posted: Mon Oct 30, 2006 12:18 pm
by oskare100
Hello,
OK, got it working now..

Best Regards
Oskar R