I get the following error code using this code:
"Warning: mysql_numrows() : supplied argument is not a valid MySQL result resource"
It's on line 55 in this php script:
"Warning: mysql_numrows() : supplied argument is not a valid MySQL result resource"
It's on line 55 in this php script:
Code:
<?php
/**
* Database.php
*
* The Database class is meant to simplify the task of accessing
* information from the website's database.
*
* Written by: Jpmaster77 a.k.a. The Grandmaster of C++ (GMC)
* Last Updated: August 17, 2004
*/
include("constants.php");
class MySQLDB
{
var $connection; //The MySQL database connection
var $num_active_users; //Number of active users viewing site
var $num_active_guests; //Number of active guests viewing site
var $num_members; //Number of signed-up users
/* Note: call getNumMembers() to access $num_members! */
/* Class constructor */
function MySQLDB(){
/* Make connection to database */
$this->connection = mysql_connect(DB_SERVER, DB_USER, DB_PASS) or die(mysql_error());
mysql_select_db(DB_NAME, $this->connection) or die(mysql_error());
/**
* Only query database to find out number of members
* when getNumMembers() is called for the first time,
* until then, default value set.
*/
$this->num_members = -1;
if(TRACK_VISITORS){
/* Calculate number of users at site */
$this->calcNumActiveUsers();
/* Calculate number of guests at site */
$this->calcNumActiveGuests();
}
}
. . . .
/**
* calcNumActiveUsers - Finds out how many active users
* are viewing site and sets class variable accordingly.
*/
function calcNumActiveUsers(){
/* Calculate number of users at site */
$q = "SELECT * FROM ".TBL_ACTIVE_USERS;
$result = mysql_query($q, $this->connection);
$this->num_active_users = mysql_numrows($result);
}
. . . .
/**
* query - Performs the given query on the database and
* returns the result, which may be false, true or a
* resource identifier.
*/
function query($query){
return mysql_query($query, $this->connection);
}
};
/* Create database connection */
$database = new MySQLDB;
?>
Comment