class within another class

Question

I am new to PHP OOP and would like to try to nest several classes within another class, to latter call them like so:

$sql = new SQL();
$sql->Head()->Description($_SESSION['page']);
  //OR
$sql->Head()->Keywords($_SESSION['page'])
  //OR
$sql->Body()->Clients($_SESSION['client'])
  //ETC
$query = $sql->Run(); // equivalent to mysql_query("...");

As you can guess, I run into some problems and ended with this poor code:

<?php
require( $_SERVER['DOCUMENT_ROOT'] . '/#some_db_directory/database.php');
//This file contains $db['host'], $db['user'], etc...

class SQL {
    public $sql;

    public function __construct() {
        global $db;
    }

    public class Head() {

        public function Description($page) {
            return "SELECT * FROM `$db['database']`.`desciption` WHERE `page` = '$page'";
        }

        public function Keywords($page) {
            return "SELECT * FROM `$db['database']`.`keywords` WHERE `page` = '$page'";
        }
    }
}

$sql = new SQL();
echo $sql->Head()->Description('home'); //For testing

Is it possible to nest classes in PHP?
If so, how is it done?

Possible duplicate of [php-method-chaining](http://stackoverflow.com/questions/3724112/php-method-chaining). — Hashem Qolami, Nov 13 '12 at 11:03
I don't think you can nest-declare classes; And even if you can, a **class** is not a **class instance**, so your `$sql->Head` is still a _class_, and `$sql->Head()->Whatever` is invalid. `$sql=new SQL()` is how you **instanize** a `SQL` class. — Passerby, Nov 13 '12 at 11:04

score 1 · Answer 1 · answered Nov 13 '12 at 11:01

What you are trying to do is called Encapsulation. Try google search on PHP encapsulation to learn more.

Here is a code example from http://www.weberdev.com/get_example.php3?ExampleID=4060,

<?php 

class App { 

     private static $_user; 

     public function User( ) { 
          if( $this->_user == null ) { 
               $this->_user = new User(); 
          } 
          return $this->_user; 
     } 

} 

class User { 

     private $_name; 

     public function __construct() { 
          $this->_name = "Joseph Crawford Jr."; 
     } 

     public function GetName() { 
          return $this->_name; 
     } 
} 

$app = new App(); 

echo $app->User()->GetName(); 
?>

I think what the OP means is PHP5 Method chaining, and adding `return $this;` at the end of each methods in class, would do the trick! — Hashem Qolami, Nov 13 '12 at 11:10

score 1 · Answer 2 · edited Jun 20 '20 at 09:12

1

I'm assuming that database.php is a database class. In that case you could do something like this.

head.php

Class Head{
    private $_db;
    private $_dbName;

    public function __construct($db, $dbName){
        $this->_db = $db;
        $this->_dbName = $dbName;
    }
    public function Description($page) {
       $results = $this->_db->query("SELECT `text` FROM `$this->_dbName`.`description` WHERE `page` = '$page'");
       return '<meta name="description" content="' . $results['text'] . '">';
    }

    public function Keywords($page) {
       $results = $this->_db->query("SELECT * FROM `$this->_dbName`.`keywords` WHERE `page` = '$page'");
       $keywords = array();
       foreach($results as $result){
           array_push($keywords, $result['word']);
       }
       return '<meta name="keywords" content="' . implode(',', $keywords) . '">';
    }
}

sql.php

require( $_SERVER['DOCUMENT_ROOT'] . '/#some_db_directory/database.php');
// Require head class file
require( $_SERVER['DOCUMENT_ROOT'] . '/#some_db_directory/head.php');   

Class SQL{
    public $Head;

    public function __construct($dbName){
        global $db;
        $this->Head = new Head($db, $dbName);
    }
}

You would then pass the name of the database into the SQL class (which propogates through to the Head class).

// Require the sql class file
require( $_SERVER['DOCUMENT_ROOT'] . '/#some_db_directory/sql.php');
$sql = new SQL('mydatabase');
echo $sql->Head->Description('home');

Again note that your database class might not return results the way I'm using them here. You will have to modify this to work with your particular database class.

edited Jun 20 '20 at 09:12

Community

1
1

answered Nov 13 '12 at 11:01

Pastor Bones

7,183
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I got lost in `head.php` @line 10, with the use of `_db->query...`. My $db array only has connection info >_< (host, user, password) . Although I would not mind creating a database class – Omar Nov 14 '12 at 08:53
@Omar ah, I was assuming $db was a database class already connected... check http://phpclasses.org they have alot of user contributed classes or what I prefer is [idiorm/paris](http://j4mie.github.com/idiormandparis/) a mysql ORM/AR. – Pastor Bones Nov 14 '12 at 10:38
My very first class file, was a database file. I wonder if I can modify it, to make it work with your example?: http://stackoverflow.com/questions/13376356/how-to-create-improve-database-php-oop-class-with-all-full-functions/13376663#comment18266724_13376663 – Omar Nov 14 '12 at 10:42
I wouldn't re-invent the wheel (except merely for educational purposes). A community developed ORM or Active Record implementation is tested and can save alot of time developing for more important things :P – Pastor Bones Nov 14 '12 at 10:44
Since I do not have a database class that would work with your example, what/how to modify your code? – Omar Nov 14 '12 at 10:46
instead of `$this->_db->query(`... would be `$this = mysql_query(..`? – Omar Nov 14 '12 at 10:48

Kami · Answer 3 · 2012-11-15T12:01:17.303

Try it like this

<?php
require( $_SERVER['DOCUMENT_ROOT'] . '/#some_db_directory/database.php');
//This file contains $db['host'], $db['user'], etc...

class SQL {
    public $sql;
    private $_head;

    public function __construct() {
        global $db;
        $_head = new HeadClass();
    }

    public function Head() {
        return $this->_head;
    }
}

class HeadClass { // Class cannot have a public access modifier

    public function Description($page) {
        return "SELECT * FROM `" . $db['database'] . "`.`desciption` WHERE page = $page";
    }

    public function Keywords($page) {
        return "SELECT * FROM `" . $db['database'] . "`.`keywords` WHERE page = $page";
    }
}

$sql = new SQL();
echo $sql->Head()->Description('home.html');
?>

I am moving the class declaration outside the class and creating an instance of the class with in SQL. This is then made available via the Head() function.

Note: For body you will need to create a separate class and use a reference in the SQL class to it like I have done for head.

what do you mean by -ve? **This is the error:** Parse error: syntax error, unexpected T_ENCAPSED_AND_WHITESPACE, expecting T_STRING or T_VARIABLE or T_NUM_STRING in /home/content/81/10038181/html/_tmp/nested-class.php on line 23 — Omar, Nov 14 '12 at 09:03
Joys of copy paste code - your original had more than one error in it. Have updated the answer with the correct code. — Kami, Nov 14 '12 at 18:49
Fatal error: Call to a member function Description() on a non-object on line 32 — Omar, Nov 15 '12 at 11:44

class within another class

3 Answers3

head.php

sql.php