我想加盟笨2個表，但它不工作

Question

我的代碼是

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該查詢返回的空列。我還添加圖片，以便更好地瞭解在此先感謝

Answer 1

function admin_profile() 
{ 
    $this->db->select('B.FNAME as customer,B.email,B.contact, C.FNAME AS 
    employee,B.active as active_user,A.active as assigned'); 
    $this->db->from('bk_ctoe as A'); 
    $this->db->join('bk_users B', 'B.ID=A.customer_id'); 
    $this->db->join('bk_users C', 'C.ID=A.employee_id'); 
return $this->db->get()->result(); 

} 

Answer 2

請嘗試在CI中的以下代碼更改您的方法，如下所示。這將返回匹配的ID與任何customer_id或employee_id

function admin_profile() 
{ 
    $this->db->select('*'); 
    $this->db->from('bk_users as A'); 
    $this->db->join('bk_ctoe as B', 'A.ID=B.customer_id OR A.ID=B.employee_id');  
    return $this->db->get()->result(); 
} 

Answer 3

嘗試使用下面的代碼爲例，

$this->db->select('*'); 
    $this->db->from('login'); 
    $this->db->join('registration', 'registration.userid = login.userid'); 
    //$query=$this->db->get('registration'); 
    $this->db->where('login.status',2); 
    $query=$this->db->get(); 
    return $query;