Re: Re: newbie question on PHP & Mysql...

From: Date: Wed, 21 Sep 2005 15:47:14 +0000
Subject: Re: Re: newbie question on PHP & Mysql...
References: 1 2 3 4 5  Groups: php.db 
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Try This $result = mysql_query("SELECT SUM(AcctInputOctets),SUM(AcctOutputOctets) FROM radacct WHERE username = '$argv[1]' "); this will make sure that even if $arg[1] is empty it still get '' (empty) as part of the query ----- Original Message ----- From: "Sylvain Gourvil" <opensource_france@yahoo.fr> To: <php-db@lists.php.net> Sent: Wednesday, September 21, 2005 3:26 PM Subject: [PHP-DB] Re: newbie question on PHP & Mysql...
Evert Meulie wrote:
Hi! I've tried your suggestions, but still get the same error message. The 'print_r($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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