无法加入代码编写器 - Unable to make join in codeigniter

Unable to make join in codeigniter

本文关键字：代码 | 更新日期: 2023-09-27

我的表结构'pages':

id int(11)
title varchar(100)
slug varchar(100)
order varchar(100)
body text
parent_id int(11)  (0/1)

我试图加入一个表到自己，这样我就可以得到有父母的页面(即parent_id与1)。我在扩展MY_Model的页面模型中使用了此方法，但它不起作用

public function get_with_parent ($id = NULL, $single = FALSE)
{
    $this->db->select('pages.*, p.slug as parent_slug, p.title as parent_title');
    $this->db->join('pages as p', 'pages.parent_id=p.id', 'left');
    return parent::get($id, $single);
}

MY_Model:

public function get($id = NULL, $single = FALSE)
{
    if($id !== NULL) {
        $filter = $this->_primary_filter;
        $id = $filter($id);
        $this->db->where($this->_primary_key, $id);
        $method = "row";
    } elseif($single) {
        $method = "row";
    } else {
        $method = "result";
    }
    if(!count($this->db->ar_order)) {
        $this->db->order_by($this->_order_by);
    }
    return $this->db->get($this->_tablename)->$method();
}

你的查询中漏掉了From

public function get_with_parent ($id = NULL, $single = FALSE)
{
   $result = $this->db->select('pages.*, p.slug as parent_slug, p.title as parent_title')
    ->from('pages');//here you need to put this code
    ->join('pages as p', 'pages.parent_id=p.id', 'left');
    return parent::get($id, $single);
}

get_with_parent($id = NULL，$ single = FALSE){

$ this -> db ->选择"页面。*,p。Slug作为parent_slug,p。标题为parent_title')

->,("页面。语言",$ this -> uri ->段(1))

->加入(页面和p, pages.parent_id = p.id ', '左');

返回父:获得(id、sengle美元);

}