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Author newbie question on PHP & Mysql...
Evert Meulie

2005-09-21, 7:55 am

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
Sylvain Gourvil

2005-09-21, 7:55 am

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

Sigrid Krug

2005-09-21, 7:55 am

Hi Evert, try:

echo mysql_result($result,0,0), "\n";
echo mysql_result($result,0,1);

You forgot a zero...

Regards, Sigrid

"Evert Meulie" <evert@witelcom.net> schrieb im Newsbeitrag
news:0tn703-4df.ln1@witelcom.com...
> 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

Sigrid Krug

2005-09-21, 7:55 am

Hi Evert, try:

echo mysql_result($result,0,0), "\n";
echo mysql_result($result,0,1);

You forgot a zero...

Regards, Sigrid

"Evert Meulie" <evert@witelcom.net> schrieb im Newsbeitrag
news:0tn703-4df.ln1@witelcom.com...
> 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

Evert Meulie

2005-09-21, 6:56 pm

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



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:
>
>
>

Sylvain Gourvil

2005-09-21, 6:56 pm

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 !



[color=darkred]
>
>
>
> Unnawut Leepaisalsuwanna wrote:
>
Evert Meulie

2005-09-22, 3:56 am

Thanks! :-)

Your solution gives me exactly what I need!


Regards,
Evert



Ziv Gabel wrote:
> 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...
>
>
>
>
>
> ****************************************
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>
> This footnote confirms that this email message has been scanned by
> PineApp Mail-SeCure for the presence of malicious code, vandals &
> computer viruses.
> ****************************************
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