Re: [PHP-DB] Re: newbie question on PHP & Mysql...

2005-09-21 Thread Ziv Gabel

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" <[EMAIL PROTECTED]>

To: 
Sent: Wednesday, September 21, 2005 3:26 PM
Subject: [PHP-DB] 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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Re: [PHP-DB] Re: newbie question on PHP & Mysql...

2005-09-21 Thread Unnawut Leepaisalsuwanna
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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