Codeigniter - 選擇id不在的位置（另一個查詢結果）

Question

我正在進入酒店預訂系統，目前我正在嘗試選擇可用的房間（未預訂）。

Rooms DB Structure:
ID
ROOM NAME
CAPACITY

HOTEL RESERVATIONS DB STRUCTURE:
ID
CHECK_IN
CHECK_OUT
ROOMS
...

這是我目前的代碼：

function searchFreeRooms($data){
  $check_in = $data['fields']['check_in'];
  $check_out = $data['fields']['check_out'];
  $this->db->select("*");
  $this->db->from('core_hotel_rooms');
  $this->db->where("id NOT IN (select rooms,total_guests from res_hotel where check_in <= '$check_in' AND check_out >= '$check_in' OR check_in <= '$check_out' AND check_out >= '$check_out' OR check_in >= '$check_in' AND check_out <= '$check_out' ) ");
  $query = $this->db->get();
  return $query->result();
 }

用戶可以一次預訂多個房間，預留房間ID存儲在“ROOMS”欄中，用逗號分隔：2,3,5

在我的前面，不應該顯示此列中存在的房間，但是我遇到了麻煩，因為只選擇了逗號之前的第一個id（房間），例如：2,3,5>僅選擇2，3,5仍然是顯示在我的面前。

問題在這里： $this->db->where("id NOT IN (select rooms,total_guests from res_hotel where check_in <= '$check_in' AND check_out >= '$check_in' OR check_in <= '$check_out' AND check_out >= '$check_out' OR check_in >= '$check_in' AND check_out <= '$check_out' ) ");

我試過這個： $this->db->where("id NOT IN (1, 2) "); 並且它完美地工作但不是上面的方法與第二個查詢。

對不起我的英語不好...

非常感謝所有能提供幫助的人！

Answer 1

最后，在你的幫助下，我解決了我的問題！

工作代碼：

function searchFreeRooms($data){
  $check_in = $data['fields']['check_in'];
  $check_out = $data['fields']['check_out'];
  $query1 = $this->db->query("select rooms from res_hotel where (check_in <= '$check_in' AND check_out >= '$check_in') OR (check_in <= '$check_out' AND check_out >= '$check_out') OR (check_in >= '$check_in' AND check_out <= '$check_out' )");
  $query1_result = $query1->result();
  $room_id= array();
  foreach($query1_result as $row){
     $room_id[] = $row->rooms;
   }
  $room = implode(",",$room_id);
  $ids = explode(",", $room);
  $this->db->select("*");
  $this->db->from('core_hotel_rooms');
  $this->db->where_not_in('id', $ids);
  $query = $this->db->get();
  return $query->result();
 }

非常感謝！

Answer 2

考慮使用括號

   $this->db->where("id NOT IN (select rooms,total_guests from res_hotel where (check_in <= '$check_in' AND check_out >= '$check_in') OR (check_in <= '$check_out' AND check_out >= '$check_out') OR (check_in >= '$check_in' AND check_out <= '$check_out' ) ) ");

Answer 3

這是您的問題的解決方案。 希望這個幫助：

function searchFreeRooms($data){
  $check_in = $data['fields']['check_in'];
  $check_out = $data['fields']['check_out'];

  $query1 = $this->db->query("select rooms from res_hotel where check_in <= '".$check_in."' AND check_out >= '".$check_in."' OR check_in <= '".$check_out."' AND check_out >= '".$check_out."' OR check_in >= '".$check_in."' AND check_out <= '".$check_out."'")->result_array();
  $room_id= array();
  foreach($query1 as $row){
     $room_id[] = $row->rooms;
   }
  $room = implode(",",$room_id);

  $query = $this->db->query("select * from rooms where id not in (".$room.")");

      return $query->result();
     }

Answer 4

$result = $this->db->select('room')
     ->get('your_table');

    foreach($result as $item) {
        $array[] = $item['id'];         
    }
    $a = implode(',', $array);

$this->where_not_in('id', $a);

做這樣的事情。