如果构造函数中的验证失败，CodeIgniter不会调用方法

Question

如果构造函数中的验证失败，我正在尝试阻止方法执行。我正在使用AJAX，我将请求发送到example/search这样的网址。我目前正在检查变量是否不是假的，但我认为有更好的方法可以做到这一点，不是吗？

class Example extends CI_Controller {

        public $error;
        function __construct() {
            parent::__construct();
                    //validation rules    
            if($this->form_validation->run() == FALSE) {
                $this->error=1;
            }

        }
        function search() {
            if(!$this->error) {
            //code
            }
        }
    }

Answer 1

<强>应用/配置/ form_validation.php

$config = array(
     'search' => array(
        array('field'=>'', 'label'=>'', 'rules'=>''),
        array('field'=>'', 'label'=>'', 'rules'=>'')
     ),
);

-

<强>控制器

public function search(){

   //is ajax request?
   if( !$this->input->is_ajax_request() )
   {
      return show_error('Bad Request!'); // bye bye, stop execution
   }

   $this->output->set_status_header('200');

   if( ! $this->form_validation->run('search') )
   {
      echo json_encode(array(
          'error' => 1,
          'errors' => array(
                'field' => form_error('field')
           )
      ));
      return; // bye bye, stop execution
   }

   //all good, continue executing
   //.....
}

修改

//For just a constructor 

protected $isValidated = TRUE;

public function __construct(){

    //check for valid ajax request
    //do this in parent class(ajax_controller) ?

    if( !$this->form_validation->run('search') )
    {
       $this->isValidated = FALSE;
       return echo json_encode( array('error' => 1));
    }

    if($isValidated)
    {
         $this->_search( $this->input->post() );
    }
}

protected function _search( $input ){}

-

class Ajax_Controller extends CI_Controller{
    public function __construct(){
       if( !$this->input->is_ajax_request() )
       {
          return show_error('Bad Request!');
       }

       $this->output->set_status_header('200');
    }
}