Hello all, I have a confusing situation on my hands, i am a member of a gaming community and we are setting up a website (yay i get to be teh techie!!) im not exactly a noob with php but i am by far not an expert, so i need some help please.
I have a table populated with mysql data, and i am using multiple submit buttons to perform actions to the selected row of information. i have 2 of the 3 buttons working great however the third is not passing on all the information that i need to pass to the next page. the only thing that is not passing is the var $name that is returned from mysql.
Below are the codes i am using. any help with how to get the var $name to pass with the $_POST information would be great, thanks in advance!
The Table: Code: (php) <?php include 'config.php'; include 'access.php';
$db = mysql_connect ($hostname, $username, $password) or die ('Failed to connect to database: ' . mysql_error()); mysql_select_db($database);
The table posts to this page: Code: (php) <?php //print_r($_POST); $submit_array = array_keys($_POST['submit']); $action = $submit_array[0]; if ($_POST) { switch ($action) { case 'edit': echo include('./includes/edit.inc'); break;
case 'delete': echo include('./includes/delete.inc'); break; case 'access': echo include('./includes/mem2admin1.inc'); break; } } ?>
and if the set button on the table is clicked the above page loads this: Code: (php) <?php if(!isset($_SESSION['clan_id'])) { die("You are not logged in and cannot view this page."); } echo '<center>'; $auth = $_POST['auth']; $name = $_POST['name']; $access = $_POST['access']; echo "