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Re: [PHP-DB] Re: newbie question on PHP & Mysql...
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| Unnawut Leepaisalsuwanna 2005-09-21, 7:55 am |
| 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:
>
>
>
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| Ziv Gabel 2005-09-21, 6:56 pm |
| Try This
$result = mysql_query("SELECT SUM(AcctInputOctets),SUM(AcctOutputOctet
s)
FROM radacct WHERE username = '$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@yahoo.fr>
To: <php-db@lists.php.net>
Sent: Wednesday, September 21, 2005 3:26 PM
Subject: [PHP-DB] Re: newbie question on PHP & Mysql...
> Evert Meulie wrote:
> 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 !
>
>
>
>
>
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