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Warning: Wrong parameter count for mysql_result() in /home/httpd/vhosts/com/

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#1 oskare100

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Posted 30 October 2006 - 05:55 PM

When I rund this part of the script:
$sql13 = "select account_username from paypal_sales where txn_id = '$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

#2 obsidian

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Posted 30 October 2006 - 06:14 PM

mysql_result() takes a minimum of 2 parameters: the result ID and the row number. You also typically need to pass the column name of the value you're after, too:
$del_account_username = mysql_result($result13, 0, 'username');

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