Re: newbie question on PHP & Mysql...

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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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