Codeigniter - 选择 id 不在的位置（另一个查询结果）

Question

我正在研究酒店预订系统，目前我正在尝试选择可用房间（未预订）。

Rooms DB Structure:
ID
ROOM NAME
CAPACITY

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

这是我当前的代码：

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();
 }

用户一次可以预订多个房间，预订的房间ID存储在“ROOMS”列中，以逗号分隔，例如：2、3、5

在我的前面，不应显示此列中存在的房间，但我遇到了麻烦，因为只选择了逗号之前的第一个 id(room)，例如：2,3,5 > 仅选择了 2 和 3， 5仍然显示在我的前面。

问题在这里：$this->db->where("id NOT IN (select rooms,total_guests from res_hotel where check_in <= '$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' ) ");

我试过这个：$this->db->where("id NOT IN (1, 2) ");，它工作得很好，但不是第二个查询的上层方法。

对不起我的英语......

非常感谢所有可以提供帮助的人！

Answer 1

终于在你们的帮助下，我解决了我的问题！

工作代码：

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();
 }

非常感谢！

Answer 2

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

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

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

做这样的事情。

Answer 3

这是您的问题的解决方案。希望对您有所帮助：

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

考虑使用括号

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