Handling Sessions - php

Question

Need some help with how to handle sessions. I am using ajax techniques to implement a group discussion platform and alot of its success depends on whether or not i can handle sessions properly, be able to see who is online etc. How can i do this efficiently. Remember, this is a typical single url ajax application where the server only responds to request. All of the form validation is done on the client side as the user enters his data. I need help with this. Below what have written so far.
<?php
include_once "../database/dbconnect.php";
session_start();
$username = isset($_POST["userNameLogin"]) ? $_POST["userNameLogin"] : $_SESSION["userNameLogin"];
$pwd = isset($_POST["passwordLogin"]) ? $_POST["passwordLogin"] : $_SESSION["passwordLogin"];
// Sending these messages to my client side validation code json-style.
if(!isset($username)){
echo("{message : 'NoName'}");
}
elseif(!isset($pwd)){
echo("{message : 'NoPW'}");
}
// creating the session variables to hold username and pwd
$_SESSION['userNameLogin'] = $username;
$_SESSION['passwordLogin'] = $pwd;
// calling the function incuded above to make connection to mysql db
dbConnection();
//query retrieves username and pwd from db and counts the result. if it is one, then they //certianly exist and if not unset the variables created above. The varibles were created
//above so i do not have to check if they exist before unsetting them.
$sQuery = "SELECT * FROM users WHERE
username = '$username' AND password = '$pwd'";
$result = mysql_query($sQuery) or die(mysql_error());
$intFound = mysql_num_rows($result);
if ($intFound == 0) {
unset($_SESSION['userNameLogin']);
unset($_SESSION['passwordLogin']);
// AD - Access Denied
echo("{message : 'AD'}");
}
else{
//a flag to set in the database who is currently online. value of 1 for users who are //online and zero for users who are not. If i want a list of those online, i check the //column called online and then check to see if the $_SESSION['username'] exist. If it //does then i know the user is online. That is what the second script is for. New to this //stuff, and do not know a better way of doing it
mysql_query("UPDATE users SET online = '1' WHERE username = '$username'") or die(mysql_error);
}
The above script should let the user login or deny access by sending messages to the validation code on client side.
As you can see, i am new to this stuff i having my share of problems. What can i do to make sure that sessions are set and unset properly i.e when user logs out.
secondly how can i monitor who is online and who is not using sessions. This is how i am trying to check who is currently online and then building a json file with the user names and sending it to the client. Json is easier to parse.
The script below tries to determine who is online
<?php
// this script determines which sessions are currently active by
// 1.) checking to see which online fields in the users table are set to 1
// 2.) by determining if a session variable has been set for these users.
// If it is not set, it means user is no longer active and script sets its online field in the users table to zero.
// After doing this, the script, then queries the users table for online fields with values one, writes them to an
// array and passes them to the client.
include_once "../database/dbconnect.php";
//include "../validation/accessControl.php";
$tempActiveUsers = array();
$activeUsers = array();
$nonActiveUsers = array();
dbConnection();
$sql = "SELECT username from users WHERE online = '1' ";
$active_result = mysql_query($sql) or die(mysql_error);
if($active_result){
while($aValues = mysql_fetch_array($active_result)){
array_push($tempActiveUsers, $aValues['username']);
}
}
forEach($tempActiveUsers as $value){
/*if($_SESSION['$value'] == $value){
$activeUsers += $value;
} */
if(isset($_SESSION['userNameLogin']) == $value){
array_push($activeUsers, $value);
}else{
array_push($nonActiveUsers, $value);
}
}
forEach($nonActiveUsers as $value1){
$sql1 = "UPDATE users SET online='0' WHERE username = '$value1'";
$set_result = mysql_query($sql1) or die(mysql_error);
}
$length = sizeof($activeUsers);
$len = 1;
$json ='{"users" : {';
$json .= '"user":[';
forEach($activeUsers as $value2){
$json .= '{';
$json .= '"username" : "' .$value2.'" }';
if($len != $length){
$json .= ',';
}
$len++;
}
$json .= ']';
$json .= '}}';
echo $json;
Please look through and give some advice. Will appreciate that very much. My project framework is up and good, but i can implement much user functionality yet because i cann't track who is online and how to manage thier sessions. If you need more background info let me know. Thanks in advance

Answer 1

Add 'Log out' button and click event handler on it which makes an ajax request to server to stop session by unsetting session vars or destroying session completely, and on ajax completion callback put a function to update user interface to show user is logged out.
Log in procedure can be done as follows: user clicks 'Log in' button and some form asking for user name and password appears. Then submit this form with ajax to a server script like your first one. Server script checks whether user name and password are valid and returns authentication information to a client via callback: failure notice upon failed login or some information about user currently logged in, e.g. user name, fullname and anything you might need about this user on client side in js. Then your client script proceeds according to login status returned from server-side script.
You should always remember about security.
Before sending any sensitive data to a client side with json, you shoud always check if session is valid and started. Client-side scripts could be easily modified and executed without your control and you should prevent undesired activity only on server side.
You should apply some escaping on user-POSTed fields before using them in sql queries to avoid sql injection attacks, e.g. by using mysql_escape_string().
And instead of building json strings, you can use json_encode() which works good for primitives, objects and arrays, you'll save some time.