Do it like this....
$sqry_linx = mysql_query("SELECT * FROM linx ORDER BY 'linkcat'");
$link = mysql_fetch_array($sqry_linx);
"Yassel Omar Izquierdo Souchay" <[EMAIL PROTECTED]> wrote in message
001c01c10eff$1235ce50$2401020a@yois">news:001c01c10eff$1235ce50$2401020a@yois...
> Hi
> I'm having
> trouble with this part of code
>
> 9: $result = mysql_db_query("users", $query);
> 10: $r=mysql_fetch_array($result);
>
> The error message is :
> Warning: Supplied argument is not a valid MySQL result resource in
> c:\inetpub\wwwroot\PhpAndMysqlTest\test2\reg1.php3 on line 9.
>
> Please if somebody knows what it's wrong
> thanks
> Yassel
>
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