Question

I am trying to learn PHP classes so I can begin coding more OOP projects. To help me learn I am building a class that uses the Rapidshare API. Here’s my class:

<?php

class RS
{
    public $baseUrl = 'http://api.rapidshare.com/cgi-bin/rsapi.cgi?sub=';

    function apiCall($params)
    {
        echo $baseUrl;
    }
}

?>

$params will contain a set of key pair values, like this:

$params = array(
    'sub'   =>  'listfiles_v1',
    'type'  =>  'prem',
    'login' =>  '746625',
    'password'  =>  'not_my_real_pass',
    'realfolder'    => '0',
    'fields'    => 'filename,downloads,size',
    );

Which will later be appended to $baseUrl to make the final request URL, but I can’t get $baseUrl to appear in my apiCall() method. I have tried the following:

var $baseUrl = 'http://api.rapidshare.com/cgi-bin/rsapi.cgi?sub=';

$baseUrl = 'http://api.rapidshare.com/cgi-bin/rsapi.cgi?sub=';

private $baseUrl = 'http://api.rapidshare.com/cgi-bin/rsapi.cgi?sub=';

And even tried $this->baseUrl = $baseUrl; in my apiCall() method, I don’t know what the hell I was thinking there though lol.

Any help is appreciated.

Answer 1

Try

class RS {
  public $baseUrl = 'http://api.rapidshare.com/cgi-bin/rsapi.cgi?sub=';

  function apiCall($params) {
    echo $this->baseUrl;
  }
}

I trust you are calling this code like so?

$rs = new RS;
$rs->apiCall($params);

Class attributes need to be prefixed with $this in PHP. The only exceptions are static methods and class constants when you use self.