Re: newbie question on PHP & Mysql...

Evert Meulie wrote:

Hi! I've tried your suggestions, but still get the same error message. The 'print_r($result);' that I added does not print anything, so that would explain why I get the errors. My idea is to call this script with a value, like: script.php value Doesn't that put the value in $argv[1] ? Regards,

    Evert

What do you use to execute your php scripts.

Php on linux ssh ? Apache ? call your script with script.php?var=value to get your value in $_GET['var'] But even if your args are emptied, it should return an error in your $result !

Unnawut Leepaisalsuwanna wrote:

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