Try This
$result = mysql_query("SELECT SUM(AcctInputOctets),SUM(AcctOutputOctets)
FROM radacct WHERE username = '$argv[1]' ");
this will make sure that even if $arg[1] is empty it still get '' (empty) as
part of the query
----- Original Message -----
From: "Sylvain Gourvil" <opensource_france@xxxxxxxx>
To: <php-db@xxxxxxxxxxxxx>
Sent: Wednesday, September 21, 2005 3:26 PM
Subject: Re: newbie question on PHP & Mysql...
Evert Meulie wrote:
Hi!
I've tried your suggestions, but still get the same error message. The
'print_r($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
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