PHP - Page Counter, Setting A Session Variable

If I have a for loop with a variable called $counter, how can I use the value of this inside a variable name. For example:

If I want a bunch of variables called $d1name, $d2name, $d3name ...etc. what's the correct syntax to use? The following gives me a string but not a variable name:

$myVariable = "$"."d".$counter."name";

Hi guys,    I would like to have a security measure in place to prevent unauthorized access to my site without a valid log on.      At the moment, it would let anyone in without destroying the session and redirecting to index page.      What would i "use" that's created in the session? what's the "best" practice    My understanding is that the session variable is stored in the browser, after a successful log in, that session variable is like baton or a key that's "passed" onto the next page.  - if someone tried to bypass the log on with the session then access is denied or redirected away.      So on my index page to start i have: 

<?php
session_start();


/* clear all session variable */
$_SESSION = array();

/* set a session variable for later use */
$_SESSION['what_page'] = "admin00";
?>

What do i need to have to use the session against unauthorized access?    my guess is: 

if(!isset($_SESSION['what_page']) || $_SESSION['what_page'] != "index.php") {
	$_SESSION = array();
	session_destroy();
	header("Location: index.php");	
	exit();
}

So to me that means;    - if 'what_page' is not set from the index page, don't go any further, re-direct (back to index)       If i remove this and use a known username and password, i am able to log into the correct page, but this session validation is the bit that's not working   please could you help?            

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

Hey guys,

Currently Im using:

$row = mysql_fetch_array($result) or die(mysql_error());
echo $row['user_family']. " - ". $row['user_registered'];
$row['user_family'] = $fam;
$_SESSION['family'] = $fam;

to take data from a mysql table & set it as SESSION family. However, I cant seem to get this to set.

The information IS being taken from mysql because its being echo'd earlier up in the code, but its just not passing to the session.

Any ideas?

Record set has 2 text fields in the form which is set in a full repeat recordset browse.  So, we get a long list of every record in the database.  However, I want to be able to click on a single record and make another page appear.  I can do this if the display is set as a table without using a text field form -- just the record variable and using a hyperlink.  But, I want to use the text field.  Wrapping the form only gets me the value of the last record displayed.  Help would be appreciated.

Hi all,

Yesterday I moved a web application from a normal cookie session to a use_trans_sid session because some users' browser didn't accept cookies.

This works great, but the session used to last for 45 minutes (set with session.gc_maxlifetime), but after the change the session times out faster.

(How) can I set the session length if I use use_trans_sid?

Thanks,

Base

PS: PHP version 5.2.13-1

My login script stores the user's login name as $_SESSION[ 'name'] on login. For some unapparent reason, i'm getting errors stating that $user  and $priv are  undefined variables, though I've attempted to define $user as being equal to $_SESSION['name'], using $user to look up the the user's privilege level (stored as the su column ) in the SQL table, and then where the result of the sql query is $priv which is then evaluated in an if statement. I can't seem to figure out why this might not be working.

The code I'm using:

<?php
session_start();
function verify() {
	//verify that the user is logged in via the login page. Session_start has already been called. 
	if (!isset($_SESSION['loggedin'])) {
	header('Location: /index.html'); 
	exit;
	}
	//if user is logged in, we then lookup necessary privleges. $_SESSION['name'] was written with the login name upon login. Privleges   // are written in db as a single-digit integer of of 0 for users, 1 for administrators, and 2 for special users. 
        $user === $_SESSION['name'];
    //Connect to Databse
	$link = mysqli_connect("127.0.0.1", "database user", "password", "database");
	if (!$link) {
  	  echo "Error: Unable to connect to MySQL." . PHP_EOL;
    	echo "Debugging errno: " . mysqli_connect_errno() . PHP_EOL;
    	echo "Debugging error: " . mysqli_connect_error() . PHP_EOL;
    	exit;
        }
        //SQL Statement to lookup privlege information. 
       if ($result = mysqli_query($link, "SELECT su FROM accounts WHERE username = $user", MYSQLI_STORE_RESULT)) {
         //LOOP TO CYCLE THROUGH SQL RESULTS AND STORE Privlege information as vairable $priv.
            while ($row = $result->fetch_assoc()) {
           $priv === $row["su"];
            }
        }
        // close SQL connection.
        mysqli_close($link);

 // Verify privleges and take action. Only a privlege of "1" is allowed to view this page. A privlege of "2" indicates special 
//accounts used in other scripts that have certain indermediate additional functions, but are not trusted administrators. 

        if ($priv !== 1) {
        echo $_SESSION['name'];
        echo "you have privlege level of $priv";
        echo "<br>";
        echo 'Your account does not have the privleges necessary to view this page';
        exit;
        }
        
}
verify();
?>

 

Hello everyone,

I can get Test 2 to successfully operate the if statement using a variable variable.

But when I try the same method using a session variable (Test 1) the if statement is not executed. Please could you tell me why the if statement in Test 1 is not being executed?

Code: [Select]

<?php
# TEST 1
$_SESSION[test_variable] = "abcd";
$session_variable_name = "_SESSION[test_variable]";
if ($$session_variable_name == "abcd")
{
echo "<br>line 373, abcd<br>";
}

# TEST 2
$test_variable = "efgh";
$test_variable_name = "test_variable";
if ($$test_variable_name == "efgh")
{
echo "<br>line 379, efgh<br>";
}

?>
Many thanks,

Stu

 v=$("#txtusername");
  $.post('../actions/check.php',{user:v.val().toLowerCase()},function(d){
  if(d=='available'){
    $("#usernamemessage").html("<span style='color:green;'>Username is available</span>");
   }else  if(d=='not-available'){
  $( "#txtusername" ).effect( "shake" );
$('#txtusername').css('border', 'solid 2px blue');
 var iserror = '1'
 alert(iserror);
}
});

alert(iserror);

The first alert displays '1', however, the second displays 'undefined'. Why is this?

Thanks

Hello. I am trying to build a PHP hit counter for each game ID in the column of my database when visited, and updating the 'fldPlays' column by +1 to show how many people visited each page. I have a huge table with a list of data, but I'll show just 3 of the columns I have. Here they a


fldID |  fldData  |  fldPlays
===================
32145    Data1            0
21543    Data2            0
75855    Data3            0
36623    Data4            0
12471    Data5            0
===================


I am trying to increase the counter number at the 'fldPlays' column to show how many users had played each game by identifying the ID first.

So for example:

User types in 'http://www.mysite.com/games.php?id=32145' in the address bar for the 'Data1' game, and then once the user enters that particular page with that ID 32145 as you see in the URL, it adds 1 to the 'fldPlays' in that row. Overall, when I echo out 'fldPlays' as text in each page, it shows how many people actually played that page.

Just to let everybody know that I am still learning PHP and I am new to counters as well.

My guess to start the PHP counter is to fetch the ID of that URL, you would have to get id like this:

Code: [Select]
<?php

if(isset($_GET['id'])) {

$conn = mysql_connect("database", "username", "password");

mysql_select_db("data");

$game_id = $_GET['id'];

$sql = "SELECT * FROM dbname WHERE fldID='$game_id' LIMIT 1";

?>
If anyone needs clarification, I will reply. Thanks for helping.

Attempting to set a variable = NOW() + INTERVAL 90 DAY;
I can run SELECT DATE_ADD(NOW(), INTERVAL 90 DAY using MYSQL, but don't know how to move result to a variable.

Currently on my website i record the statistics, for my pages they include unique views, and total page views.
at the moment every time a visitor loads a page a script updates a MySql Db.

This seems very resource intensive, is there a better way to record statistics??

 

Hi

I'm having a problem getting a query to work.

I have a simple form with user input for start and end date with format: 2009-03-19 (todays date):

$Startdate = $_POST['date'];

This works well when something is entered into the form, and afterwards using my query:

SELECT COUNT(*) as total FROM mydb WHERE Date BETWEEN '$Startdate' AND '$EndDate' ........

Problem is if user submits the form without entering anything in the date input fields, which makes sense.

I want to check if inputs has been made, and if not set af default date, but can't make it work:

if (isset($_POST['date']) && $_POST['date'] !='') {
$Startdate = $_POST['date'];}
else { $Startdate = '1980-01-01';}

How can I set $Startdate to something that can be used in the query as below doesn't work?

When I run

'select 1700-price as blah from goldclose as t2 order by dayid desc limit 1'

by itself in mysql I get a numerical result: one row, one column.

In my php script, the 1700 is actually a variable. so here it is

$changequery = sprintf("select $goldprice-price as change from goldclose order by dayid desc limit 1");
$change = mysql_query(changequery);
while ($row = mysql_fetch_array($change)) {
    printf("$row[0]"); 
}

mysql_free_result($changeresult);


I get the following error,

Warning:  mysql_fetch_array(): supplied argument is not a valid MySQL result resource in <b>/root/fuzzy/htmlmain4.php on line 99

Warning:  mysql_free_result(): supplied argument is not a valid MySQL result resource in <b>/root/fuzzy/htmlmain4.php on line 103

Not sure why? All i want is to get the result of that select statement into a variable such as $change