PHP - Multiple By Count And Session Variable
I'm not sure how to approach this but here is my problem.
I'm using a sql statement that counts the number of users in the database. I've created a session variable named $MerchantCount and then I need to multiple that number by this scheme 0-9 Needs to multiple by 2.5 10-19 Needs to multiple by 5.0 20 and up needs to multiple by 7.5 How would I do this? Thanks... Similar TutorialsHi everyone, Like the title says I need to count the amount of 1's in a column but I can't figure out how. Gr and thanx already Ryflex 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 It's late and I'm not thinking straight. I'm posting this question. Hopefully I get a reply in the morning. I have two tables. TABLE 1 - PRODUCTS TABLE 2 - LIKES I am a User who has posted these products. I want to find out ALL the Likes I have received for all my products. Here is my code. $find_products = $db->prepare("SELECT product_id FROM products WHERE user_id = :user_id"); $find_products->bindParam(':user_id', $my_user_id); $find_products->execute(); $result_products = $find_products->fetchAll(PDO::FETCH_ASSOC); if(count($result_products) > 0) { foreach($result_products as $row) { $product_id = $row['product_id']; $find_likes = $db->prepare("SELECT like_id FROM product_likes WHERE product_id = :product_id"); $find_likes->bindParam(':product_id', $product_id); $find_likes->execute(); $result_likes = $find_likes->fetchAll(PDO::FETCH_ASSOC); if(count($result_likes) > 0) { $get_likes = 0; foreach($result_likes as $row) { $get_likes++; } } } } The issue with the above code is that It only shows the Likes if I echo inside the foreach loop. And it'll show combined Likes from each of my products. But I want to actually combine ALL the Likes from ALL the products and be able show them as a single number, outside of the foreach loop. How do I do that? Edited August 17, 2019 by imgroootThis topic has been moved to MySQL Help. http://www.phpfreaks.com/forums/index.php?topic=330871.0 I insert data in mysql table row using multiple method : 1#2#3 . 1 is ID of country, 2 is Id of state, 3 is ID of town . now i have this table for real estate listings. for each list(home) i have country/state/town (1#2#3). in country table i have list of country - in country table i have list of state - in country table i have list of town. i need to The number of houses in country / state / town . my mean is : Code: [Select] USA [ 13 ] <!-- This Is equal of alabama+alaska+arizona --> ----Alabama [8] <!-- This Is equal of Adamsville+Addison+Akron --> -------Adamsville [2] -------Addison[5] -------Akron[1] ......(list of other City) ----Alaska [ 3 ] -------Avondale[3] ......(list of other City) ----Arizona [ 2 ] -------College[2] ......(list of other City) Lisintg Table : Code: [Select] ID -- NAME -- LOCATION -- DATEJOIN -- ACTIVE 1 -- TEST -- 1#2#3 -- 20110101 -- 1 2 -- TEST1 -- 1#2#3 -- 20110101 -- 1 3 -- TEST2 -- 1#3#5 -- 20110101 -- 1 4 -- TEST3 -- 1#7#6 -- 20110101 -- 1 Country Table : Code: [Select] id -- name 1 -- USA stats Table : Code: [Select] id -- countryid -- name 1 -- 1 -- albama 2 -- 1 -- alaska 3 -- 1 -- akron town Table : Code: [Select] id -- countryid -- statsid -- name 1 -- 1 -- 1 -- adamsville 2 -- 1 -- 1 -- addison 3 -- 1 -- 1 -- akron Thanks For Any Help. Say I have this array: Code: [Select] array( 0 => time 1 => time 2 => to 3 => to 4 => to 5 => fly ); and 2 variables - $occMin and $occMax as number of occurrences. If: Code: [Select] $occMin = 2; $occMax = 3; I want to produce sorted list with most occurrences at the top: Code: [Select] array( 0 => array('to', 3), 1 => array('time', 2) ); My face just melted at the thought of even beginning to tackle this. Is this somewhat possible? would it require a monster function or am I missing some native php functions which could lend a hand? Hi I am using very simple code. Here it is Code: [Select] <?php session_start(); $user = "guest"; $uid = "1"; echo $_SESSION['user']."<br />"; echo $_SESSION['uid']; ?> it displays this error Code: [Select] Notice: Undefined index: user in C:\wamp\www\DealDash\index.php on line 5 Notice: Undefined index: uid in C:\wamp\www\DealDash\index.php on line 6 how can I solve this problem? Help please Need help declaring some session variable guys. I have a login form where the member enters his 1. Pilot Callsign 2. Password I want to declare that Pilot Callsign as the session variable on authentication. Using that Pilot Callsign session variable, I will fetch data from the database relevant to his profile. I already have the whole login page coded along with the restricted access pages (not coded by me). Check this out 1. Page is coded like this and working PERFECTLY --- Code: [Select] <?php // *** Validate request to login to this site. if (!isset($_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 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 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 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! 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. hi there, i'm trying to put a message in the footer of a page which welcomes a person who is logged in with his/her name, using sessions of course; when i place this: Code: [Select] $username = $_SESSION['valid_user']; in the footer before the echo: Code: [Select] echo "You are logged in as $username"; but the session is also needed before the footer to use the username for other things , such as -checking his credit- so if place in footer the footer shows name in browser but checking credit would not happen as the assignment is at the buttom. IF i place the assignment above at the top of the file: everything works for the user and checking credit ..etc...but the footer is not there... cud not put this any clearer..sorry...hope if someone cud help...thanks 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. Hey guys, please tell me why the session var $_SESSION['return_url'] is not carrying from test.php to login.php The two echos in this file work perfectly, so i know the session var is registered... test.php <!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" xml:lang="en" lang="en"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>SWG:ANH • Bringing PreCU back to life...</title> <link rel="shortcut icon" href="favicon.ico" /> <link rel="stylesheet" type="text/css" media="screen" href="theme/main.css" /> <?php session_start(); // This section will generate a var containing the current page URL. This will be used to allow the language script to redirect users back to the page they // were on, in the language they selected. It is registered as a session variuable that will change every time the page changes. function curPageURL() { $pageURL = 'http'; if ($_SERVER["HTTPS"] == "on") {$pageURL .= "s";} $pageURL .= "://"; if ($_SERVER["SERVER_PORT"] != "80") { $pageURL .= $_SERVER["SERVER_NAME"].":".$_SERVER["SERVER_PORT"].$_SERVER["REQUEST_URI"]; } else { $pageURL .= $_SERVER["SERVER_NAME"].$_SERVER["REQUEST_URI"]; } return $pageURL; } // Here we register the URL we grabbed earlier in the session var. $return_url = curPageURL(); $_SESSION['return_url'] = $return_url; echo $_SESSION['return_url']; echo $return_url; include 'lang/enus.php'; include 'menu.php'; ?> login.php <?php session_start(); echo $_SESSION['return_url'] ; include 'dbconnect.php'; $username = mysql_real_escape_string($_POST['username']); $password = mysql_real_escape_string($_POST['password']); $return_url = $_SESSION['return_url']; echo $return_url; $result = mysql_query("SELECT COUNT(*) FROM account WHERE Username='$username' and Pass='$password'") or die(mysql_error()); $result = mysql_fetch_row($result); $result = $result[0]; if($result == 1){ $_SESSION['username'] = $username; header("Location:".$return_url); } else { echo "Wrong Username or Password. Use the back button. If you think this message is incorrect, contact the webmaster."; } ?> I know its not carrying because the echo in login.php is not returning anything and the script cant execute the header function because it has nowhere to go! Ive narrowed it down to the session var not carrying. Currently I am adding the concept of "entitlements" to my website. In the past, my "article.php" script would simply look to the URL for which article was being requested and then load it. However now that I am also adding the concept of "premium content" for "paid members", I need a way to control who sees what. What I am wondering is - from a security standpoint - how much information I should load into the $_SESSION variable. For instance, right now when a user logs in, I think I just store the "memberID" and "FirstName" and possibly "Username". It would be more efficient when a Member logs in to also retrieve their "Membership Plan" and store that in the $_SESSION variable, so that as they browse my website, each page can simply grab $_SESSION['MembershipPlan'] and run that through a function that I need to build and then determine if the user gets to access said page. However, maybe it would be more secure to have it so when a user lands on page XYZ, I would look at their "memberID" and query the database to get their "MembershipPlan"? Any thoughts on each approach? Again, my main concern is *security*, but I also suppose this plays into "performance".
Hi Guys,
Here is the code, once logged in using known credentials it should display the content "welcome..." but it doesn't, instead it is showing "you are not authorized..." as if the session['username']); isn't being taken?
<?php ini_set('display_errors',1); error_reporting(E_ALL); include_once 'includes/db_connect.php'; include_once 'includes/functions.php'; sec_session_start(); ?> <!DOCTYPE html> <html> <head> <meta charset="UTF-8"> <title>Secure Login: Protected Page</title> <link rel="stylesheet" href="styles/main.css" /> </head> <body> <?php if (login_check($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 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. 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 |