Make PHP database class generic by Generating dynamic variables

Question

Please look at the code, the array contains the table field names of the table

class User {
    public $db_fields = array('id', 'username', 'password', 'first_name', 'last_name');

    public $id;
    public $username;
    public $password;
    public $first_name;
    public $last_name;

}

The idea is to remove the public variables with a function so that it automatically creates public variable from the array which i can access ---

Example

I want to remove the

public $id;
public $username;
public $password;
public $first_name;
public $last_name;

section, and want this to be automatically generated by the $db_fields array.

So that I can access the objects by

$user = new User();
$user->username = "Ismail";

What I did was

extract($db_fields);

but it gives an error:

Parse error: syntax error, unexpected T_VARIABLE, expecting T_FUNCTION in C:\xampp\htdocs\advphp\dbclass\extractex.php on line 3

You really, *really* need to [review PHP's OOP documentation](http://php.net/language.oop5). You seem to be making a bad assumption about how class members are accessed in PHP. — Charles, Dec 15 '12 at 08:05

DiverseAndRemote.com · Answer 1 · 2012-12-15T08:37:38.377

You could remove $db_fields all together and depend on the columns returned by your SQL query like so:

class User
{
    private $_data;
    public function __construct($id = null) {
        //TODO: Load data into $this->_data for user of $id
    }

    public function __get($name) {
        if (array_key_exists($name, $this->_data))
            return $this->_data[$name];

        return NULL;
    }

    public function __set($name, $value) {
        $this->_data[$name] = $value;
    }

    public function UpdateUser(){
        //TODO: update the database with any changes to the user data or do an insert for new user
    }
}

and then you can get/set properties dynamically without needing to first declare them like so:

$newUser = new User();
$newUser->username = "Ismail";
$newUser->UpdateUser();

I understood it, but can you tell me where is the database connection or in which table the data is going to be saved. I mean, can you describe the code? — Ahmad Ismail, Jan 03 '13 at 02:23

Nerdroid · Accepted Answer · 2020-04-19T13:25:26.090

unfortunately your idea does not work if you try to use extract($db_fields); it's a method it need to run from inside a method something like a constructor or a function. extract($db_fields); it will extract the variables for you but they wont be public they will be local to that function for example if you try this

function __construct(){
    extract($db_fields);
    // the $id will be available in the constructor only
    // it will get disposed when this method finished executing
}

another approach is to use a property or setter and getter approach

<?php
class User {
    private $db_fields = array(
                        'id', 
                        'username', 
                        'password' => 'ismailPassword', 
                        'first_name',
                        'last_name'
      );

    function getValue($key){
       if (array_key_exists($key, $this->db_fields)){
            return $this->db_fields[$key];
       }
       return NULL;
    }

    function setValue($key, $value){
        $this->db_fields[$key] = $value;        
    }
}

$user = new User();
$user->setValue('username', 'Ismail');
echo " Username: ";
echo $user->getValue('username');
echo "\n\n Password: ";
echo $user->getValue('password');
?>

you can test the codes here http://codepad.org/8MwBwdut

score 0 · Answer 3 · answered Dec 15 '12 at 07:51

0

If your database result returns key value pairs you can do this:

$array = array();
while (($r = $result->fetch_array()) != false){
   $obj = new \stdClass();
   foreach($r as $key => $value){
       $obj->$key = $value;
   }
   $array[] = $obj;
}

answered Dec 15 '12 at 07:51

Ibu

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Make PHP database class generic by Generating dynamic variables

3 Answers3