Hi !
Could you do a "print_r($result)" after your mysql_query ?
Or you sure of your argv[1] ?
Sylvain Gourvil
Evert Meulie wrote:
Hi all!
I'm taking my first steps with PHP & MySQL.
Can anyone give me a hint on why this would not work?
*********************
$result = mysql_query('SELECT SUM(AcctInputOctets),
SUM(AcctOutputOctets) FROM radacct WHERE username = $argv[1] ');
echo mysql_result($result,0), "\n";
echo mysql_result($result,0,1);
*********************
I get: Warning: mysql_result(): supplied argument is not a valid MySQL
result resource
Regards,
Evert
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