|
PHP - Session Variable - Strange Thing Happening
When signed in as a user, the username shows up on the top right page. Then, if I click a particular product of another user it changes the session variable to that of the other user. Why would this happen? It's really odd.
Similar TutorialsHi. I have to check if a session is started or not, and then, check for the rank of the user. For that i use two functions like is_administrator,etc. Well, but this isnt working. I cant get any clear way to know if a session have been already started or not. Code: [Select] function is_administrator() { session_start(); if (isset($_SESSION['usr']) and empty($_SESSION['usr'])) { $l = conect(); $user = $_SESSION['usr']; $cons = "SELECT * FROM `users` WHERE usr='" . $user . '\';'; $res = mysql_query ($cons,$l); $ob = mysql_fetch_array($res); if ($ob['tipo'] == 'A') { return true; } } return false; } What i want to do is to check if a sesion exist, if not, return false. If the sesion exist, check for the user rank and return true/false. I have something strange going on with sessions set after a user logs in. After posting the form, the following code (only part of the log in script) correctly assigns the users ID to $_SESSION['id'] but after a while it starts showing up as the users level ($_SESSION['level']). I have rechecked all of my code to see if something mistakenly assigns the level to $_SESSION['id'] but everything seems fine. I am going to rewrite the log in script today, as I am sure that I can streamline it but wondered if anyone can give any advice on what it happening here to make sure I do not fall into the same trap! Thanks Steve $username = mysql_escape_string($_POST['Lname']); $password = mysql_escape_string(md5($_POST['Lpassword'])); $search = mysql_query("SELECT userid, username, password, active, level FROM users WHERE username like binary '".$username."' AND password='".$password."' AND active='1'") or die(mysql_error()); $match = mysql_num_rows($search); $level = mysql_fetch_assoc($search); if($match > 0) //The username and password is correct {//START MANUAL LOGIN & SET SESSION VARIABLES $_SESSION['user'] = $username; $_SESSION['level'] = $level['level']; $_SESSION['loggedin'] = '1'; $_SESSION['id'] = $level['userid']; 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 id not passing onto next page. when i view the source code the id value is there. <a href='view_designs2.php?id='198'> but when the next page loads there is nothing in the address bar. Code: [Select] <?php echo "<table width='1000' align='center'>"; $data=mysql_query("SELECT * FROM design WHERE jobno='$id' AND status!='rejected'") or die (mysql_error()); $counter = 0; while ($info=mysql_fetch_array($data)){ $img5=$info['name']; $id=$info['id']; $img=$info['thumbs']; echo "<td>"; echo "Design By "."$img5 "."<br><br>"; echo "<a href='view_designs2.php?id='$id'><img src='../dthumbs/$img' style='text-align: center; width:250px; max-height:250px; border: solid 10px; border-radius: 5px; border-color:tan; -moz-box-shadow: 3px 3px 4px #000; -webkit-box-shadow: 3px 3px 4px #000; box-shadow: 3px 3px 4px #000; -ms-filter: progid:DXImageTransform.Microsoft.Shadow(Strength=4, Direction=135, Color='#000000'); filter: progid:DXImageTransform.Microsoft.Shadow(Strength=4, Direction=135, Color='#000000'); ''>"; echo "<br><br></td>"; $counter = $counter + 1; if ($counter==3){ echo "<tr>"; $counter=0; } } echo "</tr></table>"; ?>' Need help declaring some session variable guys. I have a login form where the member enters his 1. Pilot Callsign 2. Password I want to declare that Pilot Callsign as the session variable on authentication. Using that Pilot Callsign session variable, I will fetch data from the database relevant to his profile. I already have the whole login page coded along with the restricted access pages (not coded by me). Check this out 1. Page is coded like this and working PERFECTLY --- Code: [Select] <?php // *** Validate request to login to this site. if (!isset($_SESSION)) { session_start(); } $loginFormAction = $_SERVER['PHP_SELF']; if (isset($_GET['accesscheck'])) { $_SESSION['PrevUrl'] = $_GET['accesscheck']; } if (isset($_POST['pilot_callsign'])) { $loginUsername=$_POST['pilot_callsign']; $password=$_POST['password']; mysql_select_db($database_brn_system, $brn_system); $LoginRS__query=sprintf("SELECT pilot_callsign, password, staff_level, firstname FROM pilots WHERE activated = 1 AND pilot_callsign=%s AND password=%s", GetSQLValueString($loginUsername, "text"), GetSQLValueString($password, "text")); $LoginRS = mysql_query($LoginRS__query, $brn_system) or die(mysql_error()); $loginFoundUser = mysql_num_rows($LoginRS); if ($loginFoundUser) { $loginStrGroup = mysql_result($LoginRS,0,'staff_level'); if (PHP_VERSION >= 5.1) {session_regenerate_id(true);} else {session_regenerate_id();} //declare two session variables and assign them $_SESSION['MM_Username'] = $loginUsername; $_SESSION['MM_UserGroup'] = $loginStrGroup; ?> --- 2. As you can see, there already is a session variable declared for Pilot Callsign But on the next page "Restricted Access Page", when I try to call this same Session Variable, it doesn't work. I tried doing this <?php echo $_SESSION['MM_Username'] ?> Moreover, I even tried to fetch data from the table like this - SELECT * FROM pilots WHERE pilot_callsign=$_SESSION['MM_Username'] Doesn't work Hi I am using very simple code. Here it is Code: [Select] <?php session_start(); $user = "guest"; $uid = "1"; echo $_SESSION['user']."<br />"; echo $_SESSION['uid']; ?> it displays this error Code: [Select] Notice: Undefined index: user in C:\wamp\www\DealDash\index.php on line 5 Notice: Undefined index: uid in C:\wamp\www\DealDash\index.php on line 6 how can I solve this problem? Help please I wonder whether someone can help me please. I'm using the script below to create a page whereby users are presented with a list of image folders they have created. Clicking on any of the folders allows the user to drill down and view the individual images. Code: [Select] <?php session_start(); $_SESSION['username']=$_POST['username']; $_SESSION['locationid']=$_POST['locationid']; ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <?php //This variable specifies relative path to the folder, where the gallery with uploaded files is located. $galleryPath = 'UploadedFiles/' . $_SESSION['username'] . '/' . $_SESSION['locationid'] . '/'; $absGalleryPath = realpath($galleryPath) . DIRECTORY_SEPARATOR; $descriptions = new DOMDocument('1.0'); $descriptions->load($absGalleryPath . 'files.xml'); $items = array(); for ($i = 0; $i < $descriptions->documentElement->childNodes->length; $i++) { $xmlFile = $descriptions->documentElement->childNodes->item($i); $path = $xmlFile->getAttribute('name'); $path = explode('/', $path); $t = &$items; for ($j = 0; $j < count($path); $j++) { if (empty($t[$path[$j]])) { $t[$path[$j]] = array(); } $t = &$t[$path[$j]]; } $t['/src/'] = $xmlFile->getAttribute('source'); $t['description'] = $xmlFile->getAttribute('description'); $t['size'] = $xmlFile->getAttribute('size'); } $basePath = empty($_GET['path']) ? '' : $_GET['path']; if ($basePath) { $basePath = explode('/', $basePath); for ($j = 0; $j < count($basePath); $j++) { $items = &$items[$basePath[$j]]; } } $files = array(); $dirs = array(); function urlpartencode(&$item, $index) { $item = rawurlencode($item); } foreach ($items as $key => $value) { if (isset($value['/src/'])) { $value['/src/'] = explode('/', $value['/src/']); array_walk($value['/src/'], 'urlpartencode'); $value['/src/'] = implode('/', $value['/src/']); $files[] = array( 'name' => $key, 'src' => $value['/src/'], 'description' => htmlentities($value['description'], ENT_COMPAT, 'UTF-8'), 'size' => htmlentities($value['size'], ENT_COMPAT, 'UTF-8') ); } else { $dirs[] = $key; } } $basePath = empty($_GET['path']) ? '' : $_GET['path']; $up = dirname($basePath); if ($up == '.') { $up = ''; } sort($files); sort($dirs); ?> <head> <title>View Image Folders</title> <meta http-equiv="Content-Type" content="text/html; charset=UTF-8" /> <link href="Styles/style.css" rel="stylesheet" type="text/css" /> <script src="Libraries/jquery/jquery-1.4.3.min.js" type="text/javascript"></script> <style type="text/css"> <!-- .style1 { font-size: 14px; margin-top: 5px; margin-right: -50px; } --> </style> <body style="font-family: Calibri; color: #505050; margin-right: 160px; margin-left: -180px;"> <div align="right" class="style1"> <a href = "index.php" /> Add Images <a/> → <a href = "javascript:document.imagefolders.submit()"> View All Images </a> </div> <form id="imagefolders" name="imagefolders" class="page" action="gallery.php" method="post" enctype="application/x-www-form-urlencoded"> <div id="container"> </div> <div id="center"> <div class="aB"> <div class="aB-B"> <?php if ('Uploaded files' != $current['title']) :?> <?php endif;?> <div class="demo"> <input name="username" type="hidden" id="username" value="IRHM73" /> <input name="locationid" type="hidden" id="locationid" value="1" /> <div class="inner"> <div class="container"> <div class="gallery"> <table class="gallery-link-table" cellpadding="0" cellspacing="0"> <thead> <tr class="head"> <th class="col-name"> Name </th> <th class="col-size"> Size </th> <th class="col-description"> Description </th> </tr> </thead> <tbody> <tr class="directory odd"> <td class="col-name"> <a href="?path=<?php echo rawurlencode($up); ?>">..</a> </td> <td class="col-size"> </td> <td class="col-description"> </td> </tr> <?php $i = 1; ?> <?php foreach ($dirs as $dir) : ?> <tr class="directory <?php $i++; echo ($i % 2 == 0 ? 'even' : 'odd'); ?>"> <td><a href="?path=<?php echo rawurlencode(($basePath ? $basePath . '/' : '') . $dir); ?>"><?php echo htmlentities($dir, ENT_COMPAT, 'UTF-8'); ?></a></td> <td>Folder</td> <td></td> </tr> <?php endforeach; ?> <?php foreach ($files as $file) : ?> <tr class="<?php $i++; echo ($i % 2 == 0 ? 'even' : 'odd'); ?>"> <td><a target="_blank" href="<?php echo $galleryPath . $file['src']; ?>"><?php echo htmlentities($file['name'], ENT_COMPAT, 'UTF-8'); ?></a></td> <td><?php echo htmlentities($file['size'], ENT_COMPAT, 'UTF-8'); ?></td> <td><?php echo htmlentities($file['description'], ENT_COMPAT, 'UTF-8'); ?></td> </tr> <?php endforeach; ?> </tbody> </table> </div> </div> </div> </div> </div> </div> </div> </div> </div> </div> </form> </body> </html> I can create the list of folders, but when I click on any of these, instead of being able to view the images, I receive the following error: Quote Warning: DOMDocument::load() [domdocument.load]: I/O warning : failed to load external entity "/homepages/2/d333603417/htdocs/development/UploadedFiles/files.xml" in /homepages/2/d333603417/htdocs/development/imagefolders.php on line 16 Warning: Invalid argument supplied for foreach() in /homepages/2/d333603417/htdocs/development/imagefolders.php on line 52 Line 16 is this line Code: [Select] $descriptions->load($absGalleryPath . 'files.xml'); and line 52 is this Code: [Select] foreach ($items as $key => $value){ However, if I change this line Code: [Select] $galleryPath = 'UploadedFiles/' . $_SESSION['username'] . '/' . $_SESSION['locationid'] . '/'; to Code: [Select] $galleryPath = 'UploadedFiles/' . 'IRHM73' . '/' . '1' . '/'; i.e. replacing the 'Session Variables' with the actual values, the page works. I've been working on this for days now, and I just can't find the solution. I just wondered whether someoen could perhaps have a look at this and let me know where I'm going wrong. Many thanks and regards hi all , i am working on a script which is oop driven and i m not much familiar with it, i appericiate if someone can help me to solve this problem , so basicaly current script is only setting one session variable to true if user login $_SESSION['is_successful_login'] , here is my code <?php include('files/db.php'); class ajaxLoginModule { private $timeout = null; private $target_element = null; private $wait_text = null; private $form_element = null; private $wait_element = null; private $notify_element = null; function __construct() { include ('config.php'); $msql = new Db; $msql->connect(); $this->is_login(); } function get_config() { $this->set_ajax_config(); } function set_ajax_config() { $this->timeout = AJAX_TIMEOUT; $this->target_element = AJAX_TARGET_ELEMENT; $this->wait_text = AJAX_WAIT_TEXT; $this->wait_element = AJAX_WAIT_ELEMENT; $this->notify_element = AJAX_NOTIFY_ELEMENT; $this->form_element = AJAX_FORM_ELEMENT; } function initLogin($arg = array()) { $this->get_config(); $this->login_script(); } function initJquery() { return "<script type='text/javascript' src='files/jquery-1.3.2.min.js'></script>"; } function login_script() { include ('files/login_script.php'); } function is_login() { if(isset($_POST['username'])) { $username = $_POST['username']; $password = $_POST['password']; $strSQL = "SELECT * FROM ".USERS_TABLE_NAME." WHERE username ='$username' AND password = '$password' "; $result = mysql_query ($strSQL); $row = mysql_fetch_row($result); /* //THIS IS WHAT I NEED $_SESSION['user'] = $row['username']; $_SESSION['id'] = $row['id']; */ $exist = count($row); if($exist >=2) { $this->jscript_location(); } else { $this->notify_show();} exit; } } function notify_show() { echo "<script>$('.".AJAX_NOTIFY_ELEMENT."').fadeIn();</script>"; } function jscript_location() { $this->set_session(); echo "<script> $('#container').fadeOut();window.location.href='".SUCCESS_LOGIN_GOTO."'</script>"; } function set_session() { session_start(); $_SESSION['is_successful_login'] = true; } } ?> i comment that line what i need is username and id to store in those session variables $_SESSION['user'] = $row['username']; $_SESSION['id'] = $row['id'] i tried to add code in function set_session but did not helped, appreciate for any help. Thanks I have a form where users enter name, username, password etc. The values are posted to a MySQL table where I also have a field called 'ID' that auto increments. I want to store that ID in a SESSION variable that I can carry over to other pages. Need help in doing this please. Hi Guys,
Here is the code, once logged in using known credentials it should display the content "welcome..." but it doesn't, instead it is showing "you are not authorized..." as if the session['username']); isn't being taken?
<?php
ini_set('display_errors',1);
error_reporting(E_ALL);
include_once 'includes/db_connect.php';
include_once 'includes/functions.php';
sec_session_start();
?>
<!DOCTYPE html>
<html>
<head>
<meta charset="UTF-8">
<title>Secure Login: Protected Page</title>
<link rel="stylesheet" href="styles/main.css" />
</head>
<body>
<?php if (login_check($mysqli) == true) : ?>
<p>Welcome <?php echo htmlentities($_SESSION['username']); ?>!</p>
<p>
This is an example protected page. To access this page, users
must be logged in. At some stage, we'll also check the role of
the user, so pages will be able to determine the type of user
authorised to access the page.
</p>
<p>Return to <a href="index.php">login page</a></p>
<?php else : ?>
<p>
<span class="error">You are not authorized to access this page.</span> Please <a href="index.php">login or register</a>.
</p>
<?php endif; ?>
</body>
</html>
I am using WAMP and have made sure the username and password is in the database correctly, how do i debug this?
the error reporting has been switched on but it doesn't help me
is the problem with:
<?php if (login_check($mysqli) == true) : ?>I am trying to follow this guide: http://www.wikihow.c...n-PHP-and-MySQL Please could i get some help on how to make the login "detect" the username from my MySQL database and display the username Thanks Attached Files login_success.php.jpg 14.31KB 0 downloads Hi all - hoping somebody can see this clearer than me... Goal: to create a very simple shopping cart page that receives $_POST info and stores it as an array in a session variable so items will be displayed in cart on clients return to shopping cart page. Problem: I can get the array to display the first item that I add to the shopping cart but each susequest item just overwrites the first because it is using the same session variable name. How do I store the next product info in a DIFFERENT session variable and display it on the shopping cart page without writing over the first? I have two pages with products on them which need to submit orders to the shopping cart page. They use a simple form with these fields: <form method="post" action="cartaction.php"> <p> <input type="submit" name="submit" value="Buy" /> <input type="hidden" name="cartaction" value="add" /> <input type="hidden" name="item" value="coppercasserole" /> </p> </form> This is what I have so far on the cartaction page which can display the first product added to the cart: <?php //If form is submitted, call the function and save all data into the array $_SESSION['form'] if($_POST['submit'] == "Buy"){ setSessionVars(); } function setSessionVars() { if ( !isset($_SESSION['cart']) ) $_SESSION['cart'] = array(); $item = array(); foreach($_POST as $fieldname => $fieldvalue) { $item[$fieldname] = $fieldvalue; } $_SESSION['cart'] = $item; echo "<table> <tr> <td>" .'<img src="images/'.$item['item'].'.jpg"' . "<td/> <td>" . $item['item'] . "</td> <td>" . '<input type="text(30)" name="value" value="1" /> <input type="submit" name="puchasednum" value="Update This One Item" />' . "</td> </tr> </table>"; } ?> I am new to PHP and have hit a brick wall as to how to display each product array as a new session variable in the shopping cart. I would like to keep it as simple as possible as I am getting easily confused! Is there a way that I can increment the session variable with something like this: <?php //session counter $counter = count($_SESSION['cart']); if (($_POST['name'] !="")){ $counter = $counter + 1; $_SESSION['cart'][$counter] = array(); } ?> Or can I pass a field as a product ID and use this somehow to create a new session variable for each product or clear the post variables somehow? Any help would be much appreciated? Thanks Hey guys, been awhile since I have been here, but I hit a little issue in reading session data reliably every time. What I am doing is reading the variables and values from a rarurlencoded string. Then putting that into a session array to populate parts of a form and fill in some hidden fields. After the form posts every now and then, about 1 in 50 or so attempts, I have one (always the same one) that just vanishes. Anyone ever experience anything like this? Hello i have problem in set session variables in .htaccess file. i have tried some codes which are belows but i did not effect in php info file. <IfModule mod_php5.c> php_value session.gc_maxlifetime 86400 php_value session.cache_expire 86400 </IfModule> please let me know if any one have solutions fot this I'm using Session variables for the first time on a site I'm developing. I had it working fine while I was doing some admin and testing in subfolders. But the problem is I'm losing the session variables when I load the page from www.example.com, but it works from www.example.com/index.php. I would be happy to post some code if needed. The $_SESSION['record_to_chage'] variable is set- I know this as I can echo it out. $openedfile = file("myfile.txt"); sort($openedfile); // foreach($openedfile as $key => $newpick) { echo "<a href=\"/editpage.php?request=$_SESSION['record_to_chgange']&newcat=$newpick\" target=\"_parent\">$newpick</a>"; echo "<br>"; } The Resulting error: Parse error: syntax error, unexpected T_ENCAPSED_AND_WHITESPACE, expecting T_STRING or T_VARIABLE or T_NUM_STRING in For the life of me I can not figure where I am going wrong. Going to get coffee and some fresh eyes. Cheers I have a session variable called $_SESSION['patchurl'] in a php file , if i get in to an else statement this session variable gets set and i go to http://yyy page. below is the snippet of the code
<?php
session_start();
?>
<?php
echo '<script type="text/javascript">' . "\n";
if(isset($_SESSION["Email"])){
echo 'window.location="http://www.xxx";';
}
else{
$_SESSION['patchurl'] = "true";
echo 'window.location="http://yyy";';
}
echo '</script>';?>
once the patchurl session variable is set i call a php file which sets an other session variable called $_SESSION["Email"]. now what happens is the $_SESSION['patchurl'] is gone and ONLY the $_SESSION["Email"] is accessible ...can i not set two session variables? why does creating a new session varible overwrites an other one even though they are called different ? am i doing something wrong ?
Edited by Ch0cu3r, 08 September 2014 - 01:05 PM. 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 page.php <a href="cart.php?action=add&id=38"> cart.php session_start(); $cart = $_SESSION['cart']; $action = $_GET['action']; switch ($action) { case 'add': if ($cart) $cart =$cart. ','.$_GET['id']; else $cart = $_GET['id']; } $_SESSION['cart'] = $cart; echo $cart; output: Insted of one time it adds the id two times. It prints : 38,38. can pls suggest me what's problem in the code. Thank's in advance. Is it good practice to put an oop object in session variable |
|||||||