Getting access to constructor variable in Codeigniter

Question

How I can get access and work on variable from constructor?

When I initialize my function hello I get error:

Message: Undefined variable: type2

Library:

class MyLibrary {


public function __construct($params)
{
    echo $params['type'];

    //I WANT TO DISPLAY TYPE2 BY HELLO FUNCTION
    $type2 = 'asdasdasd';
}

public function hello(){

    return $type2;
}
}

My controller:

  public function test()
{
    $params = array('type' => 'large', 'color' => 'red');
    $this->load->library('TwitterAPI', $params);

     //I WANT TO DISPLAY $TYPE2 
     echo $this->twitterapi->hello();

}

$this->type2 should do it with inherited methods, so public function hello(){ return $this->type2; } — ggdx
– ggdx, Commented Dec 12, 2015 at 22:27
@DanWhite Im getting error: Undefined property: MyLibrary:$type2 — Michał Lipa
– Michał Lipa, Commented Dec 12, 2015 at 22:33
Comment out / remove the echo... in the constructor. Furthermore, have you initiated the library? $this->load->library('mylibrary') — ggdx
– ggdx, Commented Dec 12, 2015 at 22:34
That echo have nothing to do with my variable $type2, and that echo works fine. Type2 of course still doesn't work with commented echo $params — Michał Lipa
– Michał Lipa, Commented Dec 12, 2015 at 22:37
***Furthermore, have you initiated the library? $this->load->library('mylibrary') *** This goes in your controller, or if you need to, autoload it in the config dir — ggdx
– ggdx, Commented Dec 12, 2015 at 22:37

Steve · Accepted Answer · 2015-12-12 22:48:15Z

2

Create a private member variable:

class MyLibrary {

    private $type2;

    public function __construct($params)
    {
        echo $params['type'];

        //I WANT TO DISPLAY TYPE2 BY HELLO FUNCTION
        $this->type2 = 'asdasdasd';
    }

    public function hello()
    {

        return $this->type2;
    }

}

answered Dec 12, 2015 at 22:48

Steve

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