Question

CodeIgniter 2.1.2

我有一个包含两个方法的类，用于此问题showone和view。后者返回小型数据库的所有项目，也可以执行搜索。另一个是针对像domain.com/showone/firstname-lastname

这样的永久链接

<?php
class Pages extends CI_Controller {

public function view($page)
    {
    //this includes a mysql search
    }

public function showone($slug)
    {
    //abbreviated version:
    $query = "SELECT * FROM mytable WHERE slug = '" . $slug . "'";
    $result = $this->db->query($query);
    if ($result->num_rows() == 0)
        {
        //here is where I'd like to use the same search that I used in showall
        }
    else
        {
        //show the one item
        }
    }

} //class
?>

因此，如果用户决定直接输入一个不会从数据库返回任何内容的URL，我想引导他搜索结果而不是显示404。

那么如何设置function searchdatabase($query)和showone同时使用的view？

Answer 1

在模型中定义该函数，加载模型，然后调用model-＆gt;方法。

<?php
//Your Model would look something like this.
class Search_Model extends CI_Model {

    public function __construct(){
      parent::__construct();
    }

    public function showone($slug)
    {
    //abbreviated version:
    //Its best to use active record for building your queries
    $this->db->where->('slug', $slug);
    $result = $this->db->get('mytable');
    if ($result->num_rows() == 0)
        {
        //here is where I'd like to use the same search that I used in showall
        }
    else
        {
        //show the one item
        }
    }

} //class

然后在你的控制器中你会这样做：

<?php
//Your Controllerwould look something like this.
class Index extends CI_Controller {
    public function __construct(){
      parent::__construct();
      //model will be loaded for each method.
      //if you're going to use this model across several controllers
      //its best to autoload it, set that in autoload.php under /app/config
      $this->load->model('search_model');
    }

    public function index(){
      $searchResults = $this->search_model->showone('slugone');
    }

更新

我刚刚意识到如果没有返回结果，你想要显示所有结果。在这种情况下，您也可以在模型中执行该逻辑。

在条件声明中，您将执行以下操作：

    if ($result->num_rows() == 0)
        {
        return $this->showall();
        }
    else
        {
        return $this->view($slug);
        }
    }

您的showall()和view()方法return $result->result();

Answer 2

您可以在控制器内使用控制器功能：

public function view($page)
    {
       $this->showone($slug);
    }

如何为多个CodeIgniter方法提供函数？

2 个答案:

更新