它可以自己加入一張表嗎？

這是我試圖加入單個數據庫表的模型函數它自己，parent :: get方法從表中獲取數據！它可以自己加入一張表嗎？

public function get_with_parents($id=NULL,$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); 
}

這是控制器！

$this->data ['pages'] = $this->Page_model->get_with_parents();

來源

2014-11-24 Cyed Safir

'左' 的左加入 – 2014-11-24 13:38:30

您也可以使用原始SQL – johnny 2017-04-17 16:54:17

假設你的單表叫pages，這應該工作 -

$this->db->select('p1.title as ptitle','p.slug as parent_slug','p.title as parent_title'); 
$this->db->from('pages AS p1'); 
$this->db->join('pages AS p','p1.parent_id=p.id','left');

來源

2014-11-24 13:40:21 hashbrown

發生數據庫錯誤錯誤編號：1066 不是唯一表/別名：'pages' SELECT'pages'。* FROM（'pages' AS pages，'pages'）LEFT JOIN'pages' as p ON'pages'.'parent_id' ='p' .'id' ORDER BY'order' 文件名：E：\ WAMP \ WWW \ CodeIgniter_2.2.0 \ SYSTEM \數據庫\ DB_driver.php 行號：330 – 2014-11-24 13:44:16

更新的代碼，然後重試 – hashbrown 2014-11-24 13:49:22

消息：未定義的屬性： stdClass :: $ parent_slug – 2014-11-24 14:18:17

public function get_with_parent($id = NULL,$sengle = FALSE) 
{ 
    $this->db->select('pages.*,p.slug as parent_slug,p.title as parent_title') 
    ->where("pages.language", $this->uri->segment(1)) 
    ->join('pages as p','pages.parent_id=p.id','left'); 
    return parent::get($id,$sengle); 
}

或

public function get_with_parent() 
{ 
    $this->db->select('pages.*, p.slug as parent_slug, p.title as parent_title') 
    ->where("pages.language", $this->uri->segment(1)) 
    ->join('pages as p', 'pages.parent_id=p.id', 'left'); 
    $query = $this->db->get('pages'); 
    return $query->result(); 
}

來源

2017-04-17 16:18:28 user2698670

它可以自己加入一張表嗎？

回答

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