CODEWITHSUNDEEP
phpmysqli
Nico Orfi
Nico Orfi

Reputation: 173

How to inherit variables to class

I am having this: 

$db_hostname = 'localhost';
$db_username = 'root';
$db_password = '';
$db_database = 'publications';

saved as connection.php but i cant inerhit it's variables in this one:

namespace form;
include_once "Form.php";
include_once "connection.php";
class login extends Form
{
    public  function showLoginfrom()
    {
    echo '<form name="loginform"  method="get">';
    $login = new Form();
    $login->AddText('username','Your username');
    $login->AddPassword('password','','Your password');
    $login->SubmitButton('Login');
    echo "</form>";
}

public function executeLogin()
{
    $connection = new \mysqli($db_hostname,$db_username,$db_password,$db_database);

    $username = ($_GET["username"]);
    $password = ($_GET["password"]);

    $query ="SELECT * FROM users WHERE username= '$username' AND password= '$password'";
    $result = $connection->query($query);

    if (mysqli_num_rows($result)== 1)
    {
        echo "Welcome";
    }
    else
    {
        echo "Login failed";
    }
}

I know that my code is unsafe, i am just experimenting. Its my first php program.

Upvotes: 1

Views: 90

Answers (2)

Reed
Reed

Reputation: 14984

Another way to do it is using a class w/ static members. Something like:

class Config {
    static public $mysqli;
    static public $dbCredentials = [];
}
include('connection.php');
Config::$dbCredentials['host'] = $db_hostname;
// and repeat for the others
//optional, extra step:
Config::$mysqli = new \mysqli(...);

Then to use, you can do:

public function executeLogin()
{
    $connection = new \mysqli(\Config::$dbCredentials['host'], ..);
    //or
    $connection = \Config::$mysqli;
    // do other stuff
}

Other notes/alternatives:

  • putting the config class in connection.php and doing that setup from there.
  • using a getter function (ie public static getConnection()) to get the mysqli connection (probably use a singleton)
  • using getter/setter functions for the credentials (ie public static getDbHost())
  • Use a proper Object Oriented approach and have executeLogin receive a mysqli object, which would be passed from the calling code.

Upvotes: 0

ʰᵈˑ
ʰᵈˑ

Reputation: 11365

Inject them into the method

For example;

public function executeLogin($db_hostname,$db_username,$db_password,$db_database)
{
    $connection = new \mysqli($db_hostname,$db_username,$db_password,$db_database);

And then call the executeLogin() method with those variables in the current scope.

Make properties

Two ways you can inject them to make properties. The first is exactly the same as above put in the __construct() method.

Another way is to include the files in the __construct() method and set them as properties. For example

private $arrDbDetails = array();

public function __construct() {
  include_once "connection.php";
  $this->arrDbDetails = array(
     'host' => $db_hostname,
     'user' => $db_username,
     'pass' => $db_password,
     'db'   => $db_database
  );

You can then access them with $this->arrDbDetails['host'], for example, in your executeLogin() method.

$connection = new \mysqli($this->arrDbDetails['host'],$this->arrDbDetails['user'],$this->arrDbDetails['pass'],$this->arrDbDetails['db']);

Using constants

You can also define the values and have them in the global scope.

define('DB_USERNAME', 'my_mysql_user');

And then call them as regular constants in your method.

Upvotes: 1

Related Questions