PHP - Session Variables Problem

Hi.  I'm brand new to this forum, so sorry if I do something wrong...

I'm someone who has thought myself to code in php to add dynamic objects to my basic static websites.  I have had no formal education.  This is not my first project however. 

I'm having problems with session variables working fine up to a point when I call a new file and then all my session variables are lost.  I have tried everything, but cannot keep them alive when I get into this or any other new file.

I display a form, get user data submitted, then display a next form and after the user submits this next form all the session variables are lost.

Anyone have any idea what is killing my session variables?

Thanks in advance.

Having some problems with PHP session i currently use cookies to log users in, but i have decided to use php session instead but i can get it to work, it simply dose not pass the variables unless include session id in every single link on my website,

can anyone tell me where i am going wrong


these are the session setting from php.in


 session.use_cookies = 1
session.use_only_cookies = 0
session.name = usersesid
session.auto_start = 1
session.cookie_lifetime = 0
session.referer_check = 1
session.cache_expire = 180
session.use_trans_sid = 1
 

Question about variable variables and using session variables for them. O.K. So if I have:

$foo = 3;
$_SESSION['bar'] = "foo";

$$_SESSION['bar']  should equal the value of $foo, however I can't get it to work. Can someone tell me what I am doing wrong..... is it formatting???

Thanks,
Thomas

I have a PHP page that offers various information from a single text file. This text file is

encrypted on the server HD.

Upon initial entry into the page, the user enters an encryption/decryption KEY and the

encrypted file is decrypted to clear text and it is available for viewing.

I have some parameters that I store in PHP session variables. I do this since various

subsequent actions by the user will require these parameters. The code is written and the whole process seems to

work well.

Since the info in these session variables is sensitive, I need to understand WHERE they are

stored. I know that it is a file on the HD, but after hours of reading the PHP Manual on

sessions, I am not finding where (HD directory) that storage is.

I have a typical shared hosting account for my web site. Mostly I want to discover is, are

the session variables in y User/file hierarchy, or are they stored in a system area where

the PHP is installed.

Whew. Sorry this was so long.

Thank you,
xphp

I am finally making the upgrade from PHP 5.3 to PHP 7.3.  When converting over the code I noticed that session some sessions are not being processed.  My test code for this is below:

 

session_start(); 
include ("../includes/connect.inc");

echo "hiiiiiiii: ".$currentDate;
echo "He  ".$_SESSION['user'];


if (isset($_SESSION['user']))
{
	$user = $_SESSION['user'];
	$currentDate = date("Y/m/d");

}
else
{
	echo "Session not started: ";
}

 

My include is listed below:

$link = mysqli_connect('localhost', 'root', ''); 
if (!$link) { 
    die('Could not connect: ' . mysql_error("Connection error message")); 
} 
mysqli_select_db($link, 'dbname') or die (mysqli_error($link)); 



date_default_timezone_set('America/Chicago');

define ("TZ_CORRECTION", 43200);

$bkcolor = "#FFFFFF";

$currentDate = (Date("Y-m-d"));

I am not getting the message "Session not started or the current date from the include".   

If it matters I am using WAMP as my testing server

Any help is greatly appreciated.

I have some difficulty identifying why the session variables' values have disappeared in my PHP code.  When I try to echo their values, nothing shows up and when I try to compare the values, it does not work.   There are no error messages to indicate something has gone wrong?   Can someone assist? Attached Files  x.txt   6.29KB   1 downloads

Hi,

I'm having a wierd problem related to session variables. It looks like variables would get duplicated inside same session, which of course should not be possible.

Background:
This problem exists on one page script which creating a quote request system where users can select different products and get quote for those. Quote system creates quote ID per customer and all products selected in one session should go under the same quote ID. Problem is that sometimes when user selects products second quote ID is created and after that products are randomly added into both of quotes. While trying to solve this problem I added simple logs to see what's going on, those are added below.

Problem seems to be caused by duplicated session variables. At somepoint it looks like session variables are not defined any more and new quote ID is created. After that the old and new session variables seems to be used randomly.

To demonstrate the problem I added session variable called 'counter' which is defined if it does not exist and incremented always when page is reloaded by submitting the form. As you can see from below logs the 'counter' variable seems to have duplicate values, see the quote value to separate two different session variable "sets".


<?php session_start();
echo "\n Current session id: ".session_id();
echo "\n _SESSION['quote']: ".$_SESSION['quote'];
$_SESSION['counter'] = isset($_SESSION['counter'])? $_SESSION['counter'] +1 : 0;
echo "\n _SESSION['counter']: ".$_SESSION['counter'];


Output when page is reloaded(form submitted):
 
 Current session id: r5i15u4s9e20ud4j6jke8ln376
 $_SESSION['quote']: 
 $_SESSION['counter']: 0
 set  _SESSION['quote']: 984
 
 Current session id: r5i15u4s9e20ud4j6jke8ln376
 $_SESSION['quote']:
 $_SESSION['counter']: 0
 set  _SESSION['quote']: 985
 
 Current session id: r5i15u4s9e20ud4j6jke8ln376
 $_SESSION['quote']: 985
 $_SESSION['counter']: 1
 
 Current session id: r5i15u4s9e20ud4j6jke8ln376 
 $_SESSION['quote']: 985 
 $_SESSION['counter']: 2 
 
 Current session id: r5i15u4s9e20ud4j6jke8ln376 
 $_SESSION['quote']: 984 
 $_SESSION['counter']: 1 
 
 Current session id: r5i15u4s9e20ud4j6jke8ln376 
 $_SESSION['quote']: 985 
 $_SESSION['counter']: 3 
 
 Current session id: r5i15u4s9e20ud4j6jke8ln376 
 $_SESSION['quote']: 984 
 $_SESSION['counter']: 2 


Could anyone explain how the 'counter' can behave such way inside same session (based on session ID)?
This is far beyond my knowledge and I would highly appreciate any advice or tip how to solve this porblem. Thanks.

BestRegards,
Laowai

I have created a test account in my database with a user level of -1 and i think my code might be wrong but i am hoping someone can spot where i have gone wrong as i cannot, also a similar problem with another session variable loggedIn this is what i get when i login this is on the index page.

Notice: Undefined index: loggedIn in C:\xampp\htdocs\Login\index.php on line 11

Notice: Undefined index: loggedIn in C:\xampp\htdocs\Login\index.php on line 17
You must be logged in to view this page!

Index page source code: 

<?php

session_start();

error_reporting(E_ALL | E_NOTICE);

ini_set('display_errors', '1');

require 'connect.php';

if($_SESSION['loggedIn'] == 1) {

	//Do Nothing

	exit();
	
} else if($_SESSION['loggedIn'] != 1) {

	echo "You must be logged in to view this page!";

	exit();
}

if($_SESSION['user_level'] == -1) {

	header("Location: banned.php");

} if(isset($_SESSION['username'])) {

	echo "<div id='welcome'> Welcome, ". $_SESSION['username'] ." <br> </div> ";

}

?>

Also if you need my login source code: 

<?php
error_reporting(E_ALL | E_NOTICE);

require 'connect.php';

session_start();

if (isset($_POST['submit'])) {
   
    $username = trim($_POST['username']);
    $password = trim($_POST['password']);
   
    if (empty($username)) {
       
        echo "You did not enter a username, Redirecting...";
       
        echo "<meta http-equiv='refresh' content='2' URL='login.php'>";
       
        exit();
       
    }
   
    if (empty($password)) {
       
        echo "You did not enter a password, Redirecting...";
       
        echo "<meta http-equiv='refresh' content='2' URL='login.php'>";
       
        exit();
       
    } 
   
    //Prevent hackers from using SQL Injection to hack into Database
    $username = mysqli_real_escape_string($con, $_POST['username']);
    $password = mysqli_real_escape_string($con, $_POST['password']);

	$result = $con->query("SELECT * FROM $tbl_name WHERE username='$username' AND password='$password'");

	$row = $result->fetch_array();

	$user_level = $row['user_level'];


// check to make sure query did execute. If it did not then trigger error use mysqli::error to see why it failed
if($result->num_rows > 0)
{

//Set default user
	$_SESSION['loggedIn'] == 1;
	$_SESSION['user_level'] == 1;
	$_SESSION['username'] == trim($_POST['username']);
	header("Location: index.php");
	exit();	
} else if($row['user_level'] == 1) {

		$_SESSION['user_level'] == 1;

		//Location admin 
		header("Location: admin.php");

		exit();

	} else if($row['user_level'] == -1) {

		$_SESSION['user_level'] == -1;

		$_SESSION['username'] == trim($_POST['username']);

		//Location banned
		header("Location: banned.php");

		exit();
	} else if($_SESSION['loggedIn'] == true) {

		//Location default user home page
		header("index.php");
	} else {

		echo "Invalid Username/Password";
	}


  //Kill unwanted session
} if(isset($_POST['killsession'])) {

	session_destroy();
	echo "<br> <br> The Session Destroyed. (Basically means you have been logged out)";
	exit();

	}

?>

I appreciate all help

Hi

Maybe a strange request and perhaps someone here knows a better way of doing what I want.

I have a sign up form on my website that automatically pulls the session var 'name' and places this in the text field for "Reference". That way I know which agent signed up a customer.

What I need to do now though is allow this variable to be injected through the url for click through banners.

This what I've done (within the header HTML Tags):

Code: [Select]
if ($_GET['Reference'] == '')
{
$_SESSION['Name']        = $my->name;
}
else
{
$_SESSION['Name']        = $_GET['Reference'];
}

However, even though this is loaded on EVERY page view the variable is empty if the user clicks another page.

Hello! I'm a bit of a novice on PHP variables still and perhaps I'm trying to run before I can walk.

I have an SQL database that I am able to successfully write to without any problems. I have also created a session variable for the user id number that I am able to retrieve on a later page. My problem is that I have hit a wall when it comes to retrieving the data from the database based upon the session variable user id.  Here is my code and error:

$_SESSION['ID_ASSIGNED']=$_POST[KSU_ID]; //this line takes the form data and successfully assigns it to the session variable on an earlier page

echo $_SESSION['ID_ASSIGNED']; //that line works normally.

Here's my problem code (the line starting with $result ). The line works fine if I input: $result = mysql_query("SELECT * FROM $usertable WHERE KSU_ID='some id number'");:

<?php

$result = mysql_query("SELECT * FROM $usertable WHERE KSU_ID=$_SESSION['ID_ASSIGNED']");

$row = mysql_fetch_array($result);

   echo "<br />Participant Number:";
   echo $row['ID'];

   echo "<br />Name: ";
   echo $row['lname'];
   echo  ",&nbsp;";
   echo $row['fname'];

   echo "<br />Student Number:";
   echo $row['KSU_ID'];

   echo "<br />Age: ";
   echo $row['age'];

   echo "<br />Gender: ";
   echo $row['gender'];

   echo "<br />Grade: ";
   echo $row['grade'];

mysql_close($con);
?>

Hi,

I am facing problems in destroying the session variables in my logout file  .I have used session_id() in my php files for ex $sid = session_id().Now when I click on logout I naviagte to my index file but the session variable $sid still remain.I have chkd it.I have kept session_start() function in a file and I am including that file in every php by require_once function.The file is palced in Includes folder..is it causing the problem?? please help me out

Also can I use session_start() more than once in a single php file?

My logout file is :

<?php
        session_start();
   session_destroy();
   header("location:index.php");
   exit();
?>

Hello everyone and thanks in advance for the help. I am working on a permission system for a site I am creating that will be used to restrict access to various areas of the site. I have a table with all the permission that is linked to a group that the user is a member of. I then use a query to get a list of the permission ids the user is part of and put those into the session array. I used a basic login script I found online I planned on modifying to add my own features and to make more secure.

Originally it had just a member id and member name variable in the session array saved in the login script which are both working fine after login. I am using a dynamic variable to set one session variable for each permission with the id being the changing part in the variable name. The problem I found is that the new variables are not being set. When i removed the redirect from my login script so I could check the problem I found that the temp variable, which is also set before it saves to the session array, is blank as well. If I hit refresh on my browser the new variables become set and all is well.

I was talking to nvee in irc chat who said I need to make a function on the page I am redirected to that checks if the variable exists. I looked through the original code of the login script I am using and found that this seems to be how they did it as well so I added an if statement to my index page to check if the variable has a value with and without using if isset. In both cases the new variable is still not getting set. He also gave me some pointers to help make my login checks more secure but the main thing IM concerned with right now is to make my new permission variables accessible to the session. My login script is available in my pastebin.

Hello everyone, I'm sitting here in a position where I can't work on my site or test any code, but my mind is racing about what I can do to solve a particular problem that I have.  I won't get into the problem because it would take more time to type than I have right now, but I have an idea for a simple solution, just not sure if it will work or not.  So, here's my question: Is it possible to have multiple session variables during a session?

Such as:
$email=$_SESSION['email'];
$user=$_SESSION['userid'];

If this is possible, my problem is solved (I think)... otherwise, I have to keep thinking about it.  Thanks for any help!

Is there a way to use extract to make session variables? If so how would you go about doing it?

Hi All! I've written up a script for my website. It\ is basically a virtual job quest. My queries are all correct it just isn't registering the variable for the session. It is $-SESSION[theid']. I want to be bale to use it in my table but I get an error. How do I write this in my SQL query for it to work. The page (when no errors), doesn't show my data. Here is my ocde:

Code: [Select]
<?php
session_start();
include("config536.php");
?>
<html>
<head>
<link rel="stylesheet" type="text/css" href="style.css" />
</head>

<?php
if(!isset($_SESSION['username'])) {
echo "<ubar><a href=login.php>Login</a> or <a href=register.php>Register</a></ubar><content><center><font size=6>Error!</font><br><br>You are not Logged In! Please <a href=login.php>Login</a> or <a href=register.php>Register</a> to Continue!</center></content><content><center><font size=6>Messages</font><br><br></center></content>";
}

if(isset($_SESSION['username'])) {
echo "<nav>$shownavbar</nav><ubar><img src=/images/layout/player.gif><a href=status.php>$showusername</a>.......................<img src=/images/layout/coin.gif> $scredits</ubar><content><center><font size=6>Basic Quests</font><br><br>";

$startjob = $_POST['submit'];
$jobq = "SELECT * FROM jobs WHERE username='$showusername'";
$job = mysql_query($jobq);
$jobnr = mysql_num_rows($job);

if($jobnr == "0") {
?>
<form action="<?php echo "$PHP_SELF"; ?>" method="POST">
<input type="submit" name="submit" value="Start Job"></form>
<?php
}
if(isset($startjob)) {
$initemidq = "SELECT * FROM items ORDER BY RAND() LIMIT 1";
$initemid = mysql_query($initemidq);
while($ir = mysql_fetch_array($initemid)) {
$ids = $ir['itemid'];
}
mysql_query("INSERT INTO jobs (username, item, time, completed) VALUES ('$showusername', '$ids', 'None', 'No')");
$wegq = "SELECT * FROM items WHERE itemid='$ids'";
$weg = mysql_query($wegq);
while($wg = mysql_fetch_array($weg)) {
$im = $wg['image'];
$nm = $wg['name'];
$id = $wg['itemid'];
}
$_SESSION['theid'] = $id;

echo "<font color=green>Success! You have started this Job!</font><br><br>Please bring me this item: <b>$nm</b><br><br><img src=/images/items/$im><br><br><br>";
echo $_SESSION['theid'];
}

if($jobnr == "1") {
$finish = $_POST['finish'];
$okgq = "SELECT * FROM items WHERE itemid='$yes'";
$ok = mysql_query($okgq);
while($ya = mysql_fetch_array($ok)) {
$okname = $ya['name'];
$okid = $ya['itemid'];
$okimage = $ya['image'];
}
echo "Where is my <b>$okname</b>?<br><br><img src=/images/items/$okimage><br><br><br>";
echo $_SESSION['theid'];
?>
<form action="<?php echo "$PHP_SELF"; ?>" method="POST">
<input type="submit" name="finish" value="I have the Item"></form>
<?php
}
}

if(isset($finish)) {
$cinq = "SELECT * FROM uitems WHERE theitemid='$_SESSION[theid]'";
$cin = mysql_query($cinq);
$connr = mysql_num_rows($cin);
if($connr != "0") {
echo "<font color=green>Success! You have the item.</font>";
} else {
echo "<font color=red>Error! You do not have my item!</font>";
}

}
?>
. I basically just want to know how I can set this session as a variable. Also..I have a user login on every page and I want to be able to destroy JUST THE "theid" session and NOT the username session. How would I do that too? thanks for the help in advance!