Re: Re: newbie question on PHP & Mysql...

Hi, I guess you used a single quote over the query so the text, $argv[1], was entered into the query rather than the value inside it. try: $result = mysql_query('SELECT SUM(AcctInputOctets), SUM(AcctOutputOctets) FROM radacct WHERE username = ' .$argv[1] ); OR $result = mysql_query("SELECT SUM(AcctInputOctets), SUM(AcctOutputOctets) FROM radacct WHERE username = $argv[1]"); should do the trick 21nu Sylvain Gourvil wrote: > Hi ! > > Could you do a "print_r($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 > >