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Another Login System Question?


JeffC
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Hi,I have finished making the multiple user account interface, but, now I am having trouble with another feature of the interface.The purpose of the code that I have added, is to reference the "shared" table of the database. But when I try to call a cell from the table farther below, I get the following error:Notice: Undefined index: shared in [filepath] on line 134. What I am currently trying to do to call the second table and then store it as a variable, is this:

$query = "  SELECT * FROM shared  ";$_data = ['shared'];

The error on line 134 is caused by the following code:

<?php echo htmlentities($_data['shared']['variable'], ENT_QUOTES, 'UTF-8'); ?>

The full relevant code is below.

<!doctype html><?php	require("common.php");	$submitted_username = '';$query = "  SELECT * FROM shared  ";$_data = ['shared'];	if(!empty($_POST))	{		$query = "			SELECT				id,				username,				password,				salt,				email,	firstname,	lastname,	accountType			FROM users			WHERE				username = :username		";		$query_params = array(			':username' => $_POST['username']		);				try		{			$stmt = $db->prepare($query);			$result = $stmt->execute($query_params);		}		catch(PDOException $ex)		{ .			die("Failed to run query: " . $ex->getMessage());		}				$login_ok = false;				$row = $stmt->fetch();		if($row)		{			$check_password = hash('sha256', $_POST['password'] . $row['salt']);			for($round = 0; $round < 65536; $round++)			{				$check_password = hash('sha256', $check_password . $row['salt']);			}						if($check_password === $row['password'])			{				$login_ok = true;			}		}				if($login_ok)		{			unset($row['salt']);			unset($row['password']);			$_SESSION['user'] = $row;						header("Location: private.php");			die("Redirecting to: private.php");		}		else		{			print("Login Failed.");  			$submitted_username = htmlentities($_POST['username'], ENT_QUOTES, 'UTF-8');		}	}  ?><body data-accent="blue">  <!-- Prompt IE 6 users to install Chrome Frame. Remove this if you support IE 6.	   chromium.org/developers/how-tos/chrome-frame-getting-started -->  <!--[if lt IE 7]><p class=chromeframe>Your browser is <em>ancient!</em> <a href="http://browsehappy.com/">Upgrade to a different browser</a> or <a href="http://www.google.com/chromeframe/?redirect=true">install Google Chrome Frame</a> to experience this site.</p><![endif]-->   <header id="nav-bar" class="container-fluid">	  <div class="row-fluid">		 <div class="span8">			<div id="header-container">			   <div class="dropdown">				  <a class="header-dropdown dropdown-toggle accent-color" data-toggle="dropdown" href="#">					 Cloud<?php echo htmlentities($_data['shared']['variable'], ENT_QUOTES, 'UTF-8'); ?>				  </a>			</div>			</div>		 </div>   </div>   </header>   <div class="container-fluid">	  <div class="row-fluid">		 <div style="text-align: center;" class="metro span12">	<p>Woah there champ! Before you move any further, you need to be logged in!</p>	<form action="login.php" method="post">  Username:<br />  <input type="text" name="username" value="<?php echo $submitted_username; ?>" />  <br /><br />  Password:<br />  <input type="password" name="password" value="" />  <br /><br />	   <input type="submit" class="btn btn-primary" value="Log Me In!">	</form>

If someone could help that would be great!

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After you do this:

$query = "SELECT * FROM shared";$_data = ['shared'];

Assuming you're using a recent version of PHP, then $_data will be an array with the string "shared" in the first index: $_data = array(0 => 'shared'); That's why $_data['shared'] is undefined. If you want to make a sub-array with the key 'shared', then it looks like this: $_data = array('shared' => array()); That being said, if you expect $_data to hold data from the database, then you also need to execute the query and populate $_data with the result. You define that variable called $query but then you don't do anything with it, including actually running it.

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After you do this:
$query = "SELECT * FROM shared";$_data = ['shared'];

Assuming you're using a recent version of PHP, then $_data will be an array with the string "shared" in the first index: $_data = array(0 => 'shared'); That's why $_data['shared'] is undefined. If you want to make a sub-array with the key 'shared', then it looks like this: $_data = array('shared' => array()); That being said, if you expect $_data to hold data from the database, then you also need to execute the query and populate $_data with the result. You define that variable called $query but then you don't do anything with it, including actually running it.

So what I should do, is switch $query, with $_data, and then, after$_data = "SELECT * FROM shared";$_data = ['shared'];, I need to add $_data = array('shared' => array()); below this?
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It doesn't matter what you name the variable that holds the query, but if you want to get data from the database then you need to actually run the query. Defining a variable that contains a query isn't enough to interact with a database, you need to actually connect to the database and send the query to it. You're executing another query on the page to get the user information, that's how to execute a query. Just defining a variable that contains a query doesn't execute the query on a database. After you run the query then you can get the data and build an array to contain whatever data you want from the database.

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