代码点火器将库实例从一个函数传递给下一个函数

Question

从代码点火器函数我正在加载一个带有参数的库，它执行一堆事情并初始化库实例。 将此实例传递给另一个函数的正确方法是什么？

编辑：

public function init()
{        
         //...
         $dataArr = array('directory' => $myDir, 'run' => TRUE);
     $this->load->library('lib_mylib', $dataArr);           


         // here I want to pass the lib_mylib instance to final();
                 // lib my lib has variables that gets set when loaded, I don't want to reloaded it again in the next function because it performs some operations that should only happen once. How can I get a handle of that initiliazed library inside final?
    }        

public function final($inputLibMyLib)
{           


}

Answer 1

假设此代码在Controller中，只需在Controller的构造函数中加载库...

class Example extends CI_Controller {

    // load your Controller wide library here
    public function __construct()
    {
        parent::__construct();

        $dataArr = array('directory' => $myDir, 'run' => TRUE);
        $this->load->library('lib_mylib', $dataArr);
    }

    public funciton init()
    {
        $this->lib_mylib->function(); // is available
    }

    public funciton final()
    {
        $this->lib_mylib->function(); // is also available
    }

}

每次请求函数时都会调用此构造函数。

的澄清 的

在构造函数中执行此操作是最佳选择，因为它可以保证在以下任何函数之前加载它。

...然而

如果您按顺序调用函数，则加载的库将保持...

例如

class Example extends CI_Controller {

    // load your Controller wide library here
    public function __construct()
    {
        parent::__construct()
    }

    public funciton init()
    {
        $dataArr = array('directory' => $myDir, 'run' => TRUE);
        $this->load->library('lib_mylib', $dataArr);

        $this->lib_mylib->function(); // is available

        // call next function
        $this->final();
    }

    public funciton final()
    {
        // is also available if final() is called from within init()
        $this->lib_mylib->function(); 
    }

}

Answer 2

只需确保将库实例by reference传递给您的函数，它应该可以正常工作。