显示错误的联接表的Codeigniter 4分页

Question

我在做代码点火器项目。我显示的新闻表格与加入左新闻类别和分页。当我尝试下面的代码时，我得到了一个错误Call to undefined method CodeIgniter\Database\MySQLi\Builder::paginate()。如何使用连接和分页显示新闻数据？

    $db      = \Config\Database::connect();
    $news_tbl = $db->table('tbl_news')->join('tbl_category', 'tbl_news.category_id = tbl_category.category_id');
    $data['news_fetched'] = $news_tbl->paginate(10);
    $data['pager'] = $news_fetched->pager;
    $data['links'] = $data['pager']->links();

Answer 1

$db->table()创建一个查询生成器对象。分页适用于模型对象。你的分页方式不对。您应该创建一个名为tblNewsModel的模型文件，它将为您处理分页。

<?php
namespace App\Models;

class tblNewsModel extends \CodeIgniter\Model {

    protected $table = 'tbl_news';
    protected $primaryKey = 'your_pk_id';

    // your function to paginate
    public function paginateNews(int $nb_page) {
        return $this->select()->join('tbl_category', 'tbl_news.category_id = tbl_category.category_id')->paginate($nb_page);
    }

}

然后在您的控制器中，只需创建该模型的一个新实例，并要求他给您分页。

$tblNewsModel = new \App\Models\tblNewsModel();
$data['news_fetched'] = $tblNewsModel->paginateNews(10);
$data['pager'] = $tblNewsModel->pager;
$data['links'] = $data['pager']->links();