Using the resources from one class in another (interface? extend?)

[amnesia623]

Hello,

So I have been trying to work more and more in OOP so I can get as used to it as I can.

One thing that isn't clear is how to build one class so it uses the results/resource of another. For example, say I have one class that connects to a database:

class db {
	function __construct() {
	mysql_connect("localhost", "user", "pass") or die(mysql_error());
	mysql_select_db("database") or die(mysql_error());

$result = mysql_query("SELECT * FROM Items")
or die(mysql_error());  
}
}

(in the future I would like to pass the column names to the class instead of using the *)

Then I want to use the results from the above class to be used for this:

class listing extends db {
function list_results($this->result) {
	$counter = 1;
	echo "<ul>";
		while($row = mysql_fetch_array($this->result))  {
			if ($counter == 1) {
				$row_style = "odd";
				$counter = 2;
			} else {
				$row_style = "even";
				$counter = 1;
			}
			echo "<li class=\"$row_style\"> <ul> \n" ;
			echo "<li>" . $row["ItemTitle"] . "</li>";
			echo "</li> </ul>";

	}
	echo "</ul>";
}
}

by creating the instance:

$output = new listing;
$output->list_results()

what am I missing?

Do I need an interface or do I extend like above?

Thanks

[laserlight]

amnesia623 wrote:

One thing that isn't clear is how to build one class so it uses the results/resource of another.

It depends on how the classes are related. In your example, you used inheritance, so you are saying that a listing is-a db.

But what is a listing? Is it supposed to be a database class? It looks like it is supposed to model some kind of item listing, so making it inherit from db is not correct. More likely you would either pass a reference to a db object to the constructor of listing, or you would pass a reference to a db object to the relevant methods of listing.

Oh, and instead of writing your own database wrapper class, I recommend that you use the PDO extension instead.

[amnesia623]

ok -

Seems that I need some more thought on this (and research). A listing is supposed to be a list of items pulled from the database.

Thanks for the info on the PDO extension. Is this a standard extension? The reason I ask is because my host doesn't have the mysqli extension installed or else I would have used that.

[laserlight]

amnesia623 wrote:

A listing is supposed to be a list of items pulled from the database.

Instead of concentrating on the "pulled from the database", concentrate on the list of items itself. That is, what kind of list is it? What is the interface that you would want to define for the list? What kind of items will be stored?

amnesia623 wrote:

Thanks for the info on the PDO extension. Is this a standard extension? The reason I ask is because my host doesn't have the mysqli extension installed or else I would have used that.

It is "standard" in the sense that it is typically bundled with PHP, but it may or may not be enabled (though these days it usually is).

[amnesia623]

So what would you do if your host has PDO disabled? (per phpinfo)

[laserlight]

amnesia623 wrote:

So what would you do if your host has PDO disabled? (per phpinfo)

Request that the host enable it, and move to another host if the host refuses. There is simply no excuse to not provide PDO to your customers these days.

[NogDog]

amnesia623;10907379 wrote:

So what would you do if your host has PDO disabled? (per phpinfo)

• Ask the host to enable it?

• Switch hosts?

• Grab a 3rd-party DB abstraction class that you like and upload it to your site and use it? (See PEAR.php.net, sourcefoge.com, etc.)

• Write your own?

😉

PS: Is your host at least running PHP 5.2.x? If not, I'd go immediately for the "Swich hosts" option.

[amnesia623]

it's 5.2.8 that is being run.

[NogDog]

amnesia623;10907384 wrote:

it's 5.2.8 that is being run.

Seems like they really ought to at least have MySQLi enabled since that's now the preferred MySQL-specific interface.

[amnesia623]

Nope. You would think tho...

msqli doesn't work either. I am just trying out this new host. So far...well...

[NogDog]

Anyway, back to your original question: you can instantiate whatever DB class you end up using, then pass it as a parameter to the constructor of any class that needs to utilize it. For example:

<?php

class MyMySQL
{
   private $dbLink;
   public $lastError;

   public function __construct($host, $user, $pwd, $dbName)
   {
      $this->dbLink = mysql_connect($host, $user, $pwd);
      if($this->dbLink === false)
      {
         throw new Exception("Connect failed: " . mysql_error());
      }
      if(mysql_select_db($dbName, $this->dbLink) === false)
      {
         throw new Exception("Select DB failed: " . mysql_error());
      }
   }

   public function query($sql)
   {
      $result = mysql_query($sql, $this->dbLink);
      if($result == false)
      {
         $this->lastError = mysql_error();
         error_log($this->lastError);
      }
      return $result;
   }
}

class Example
{
   private $db;

   public function __construct(MyMySQL $db)
   {
      $this->db = $db;
   }

   public function test()
   {
      $result = $this->db->query('SELECT * FROM some_table');
      return $result;
   }
}

try
{
   $db = new MyMySQL('localhost', 'user', 'password', 'test_db');
   $obj = new Example($db);
   $result = $obj->test();
}
catch(Exception $e)
{
   error_log($e);
   echo "<p class='error'>Oops!</p>";
}