php - 如何获取表2中没有的值

Question

如何获取table_1中不在table_2中的id
我的 table 看起来像这种类型。
表_1

---------------------------------------------
|  id      |    user                        |
---------------------------------------------
|   1      |    Jack                        |
---------------------------------------------
|   2      |    John                        |
---------------------------------------------

表_2

------------------------------------------
|  web_id      |    website              |
------------------------------------------
|   2          |   Facebook              |
------------------------------------------
|   3          |   Google+               |
------------------------------------------

我想要codeigniter查询

$this->db->select("*");
        $this->db->from("table_1");
        $this->db->where_not_in('id', "table_2.web_id");
        return $this->db->get();

Answer 1

试试这个

 SELECT id
 FROM table_1
 WHERE id NOT IN
 (
       SELECT web_id FROM table_2 WHERE 1
 );

使用 CodeIgniter Active Records 查询如下

  $this->db->select('*');
  $this->db->where('id NOT IN (SELECT web_id FROM table_2 WHERE 1)', NULL, FALSE);
  $query = $this->db->get('table_1');