Calling a class method inside another class

[play_]

How can it be done?

say i have

 

class SQL {
  function query($query) {
$this->result = @mysql_query($query, $this->linkid);
if(! $this->result) { echo '<br />Query failed'; }
$this->querycount++;
return $this->result;
   }
}



class B {
  function hello() {
     // need to use function 'query' here.
     // Do i create an instance of SQL in here and use it like $sql->query("select * from users")... ?
  }
}

[btherl]

For that example, yes you must create an instance of SQL, because the method refers to $this.

 

But in general, you can just use SQL::query($var);

[play_]

Yes, i can.

Wasn't working at first, but now it is.

 

 

class player {
    function retrieveStats($userID) {
$sql = new mysql();
$sql->query("SELECT * FROM user_character WHERE userID = '$userID'");
$this->result = $sql->fetchArray();
return $this->result;
    }
}

class mysql {
    function query($query) {
$this->result = @mysql_query($query, $this->linkid);
if(! $this->result) { echo '<br />Query failed'; }
$this->querycount++;
return $this->result;
    }
}

 

Use it like this:

$player = new player();

$row = $player->retrieveStats('1'); // outputs 1. 1 was the playerID sent as paremeter

 

[play_]

For that example, yes you must create an instance of SQL, because the method refers to $this.

 

But in general, you can just use SQL::query($var);

 

never came upon :: before so i'll stick to what i just did. =]

Thanks though.

[play_]

Howcome i can't initiate an instance of mysql class in the __construct() method?

	function __construct() {
	$sql = new mysql();
}

 

 

 

gives:

 

Notice: Undefined variable: sql in local/path on line 209

[Jenk]

The double colon is a static operator, you cannot use it to call an instantiated object, it will call the method of the class as if it were just a function.

 

Learn to pass the objects around, in true to OOP.

 

<?php

$mysql = new SQL;
$B = new B($mysql); // pass the $mysql object to $B so it can store it within it's properties and utilise it.

?>