Using Classes In Php

[bigjoe11a]

What I'm trying to do is access a class to access a mysql database and do an query, The problem is that the class as it is now needs to get the settings from a config.php file and well that doesn't seem to work. if I add the config.php file to the login.php script. It tells me I have a 2nd session started and values are all ready set. So I had to remove it from the top part of my login script.Can some one please help. The Class is blow

public function verifyDatabase()  {    //Database Connection Data    mysql_connect($settings['db_host'], $settings['db_user'], $settings['db_pass']) or die(mysql_error());    mysql_select_db($settings['db_name']) or die(mysql_error());    $data = mysql_query("SELECT ID FROM ".$settings['db_prefix']."users WHERE user_name = '{$this->_username}' AND user_pass = '{$this->_passmd5}'");    if(mysql_num_rows($data))	  {	    list($this->_id) = @array_values(mysql_fetch_assoc($data));	    return true;	  }    else	  { return false; }  }

The log in php code

<?phpif(isset($_POST['login'])){  include('sources/class.login.php');  $login = new Login();  if($login->isLoggedIn()) {	  define('LOG_IN',true);	 //header('location: index.php');  }  else    $login->showErrors();}$token = $_SESSION['token'] = md5(uniqid(mt_rand(),true));?><form method="POST" action="<?php echo $_SERVER['PHP_SELF'];?>"><table> <tr><td>Username:</td><td><input type="text" name="username" /></td></tr> <tr><td>Password:</td><td><input type="password" name="password" /></td></tr></table><input type="hidden" name="token" value="<?php echo $token;?>" /><input type="submit" name="login" value="Log In" /></form>

[boen_robot]

Consider including your configuration data earlier and with constants instead of variables. That way, you can just add

include 'config.php';

at the top of your main PHP file ("index.php" I'd assume) and then use the constants within your function. An alternative - IMHO better - approach is to use what's commonly called "dependency injection" i.e. pass the configuration data to the class/method/whatever that needs it. Since logging in is necessarily done against a database, you might inject the DB configuration at the Login() class constructor.

[bigjoe11a]

Thanks, I tried that I still have the same problem. It only seems to be in the _DbConnect(); and I don't have a constructor. for this class. DataManager. I do how ever for the login class I do. a constructor for that. That's below and I still have the same problem, The class for the DataManager is below all so...

public function __construct()  {    $this->_errors = array();    $this->_login  = isset($_POST['login'])? 1 : 0;    $this->_access = 0;    $this->_token  = $_POST['token'];    $this->_id	   = 0;    $this->_username = ($this->_login)? $this->filter($_POST['username']) : $_SESSION[SESS_USER];    $this->_password = ($this->_login)? $this->filter($_POST['password']) : '';    $this->_passmd5  = ($this->_login)? md5($this->_password) : $_SESSION['password'];	      }

Class DataManager()

class DataManager{        	 private static function _DbConnect()    {	    	    mysql_connect($settings['db_host'],$settings['db_user'],$settings['db_pass']) or die(mysql_error());	    mysql_select_db($settings['db_name']) or die(mysql_error());    }        public static function GetTotoalUsers()    {	    self::_DbConnect();	    	    list($count) = $array_values(mysql_fetch_assoc(mysql_query("SELECT count(*) FROM ".$settings['db_prefix']."users")));	    return $count;    }        public static function GetProfile()    {	    	    self::_DbConnect();	    	    $profile = mysql_query("SELECT * from " . $settings['db_prefix'] . " profile where id=".SESS_USERID."LIMIT = 1");	    	    $row = mysql_fetch_assoc($profile);		    $list = $row;		    	    return $list;    }    }

Any way as soon as I try and connect to the mysql database from any one of these classes. I get the same error. The values $settings['db_host']; is not fount like the others. Can you give me an idea on what to do, Sorry. I never have done this before. This is all so new to me. Joe

[justsomeguy]

The problem is that the $settings variable does not exist inside the function. The three options are to either use a global variable, which is the least desirable, define the settings as constants instead of variables, or passing the $settings variable or individual values into the constructor. Since multiple methods in the class all use those values, it makes sense to pass them to the constructor.

[bigjoe11a]

define the settings as constants instead of variables, Thanks, That's what I had to do and it's working, Maybe I should start using them more. Most of the time nothing else works.Thank YouJoe