php - 不同/连接两个表

Question

商店表:

   +--+-------+--------+
    |id|name   |date    |
    +--+-------+--------+
    |1 |x      |March 10|
    +--+-------+--------+
    |2 |y      |March 10|
    +--+-------+--------+

分类表:

+--+-------+
|id|title  |
+--+-------+
|1 |tools  |
+--+-------+
|2 |foods  |
+--+-------+

商店类别表 (shop_cats):

+--+-------+--------+
|id|shop_id|cat_id  |
+--+-------+--------+
|1 |1      |1       |
+--+-------+--------+
|2 |1      |2       |
+--+-------+--------+

我想按类别获取商店(类别存储在 $cat 数组中)

     $this->db->select('shops.*');
     $this->db->from('shops');
     if(!empty($cat))
     {
         $this->db->join('shop_cats' , 'shop_cats.shop_id = shops.id' );
         $this->db->where_in('shop_cats.cat_id' , $cat);
     }


    $this->db->limit($limit , $offset);
    $res = $this->db->get();

我的问题是它返回重复的结果例如在这张表中

+--+-------+--------+
|id|shop_id|cat_id  |
+--+-------+--------+
|1 |1      |1       |
+--+-------+--------+
|2 |1      |2       |
+--+-------+--------+

如果我想要 (1,2) 类别的商店，我会得到 id = 1 的商店，两次。我希望它只返回每个商店一次，没有任何重复。

我试过使用分组方式

 if(!empty($cat))
         {
             $this->db->join('shop_cats' , 'shop_cats.shop_id = shops.id' );
             $this->db->group_by('shop_cats.shop_id');
             $this->db->where_in('shop_cats.cat_id' , $cat);
         }

没用，我也试过了

 if(!empty($cat))
       {         $this->db->select('DISTINCT shop_cats.shop_id');
             $this->db->join('shop_cats' , 'shop_cats.shop_id = shops.id' );
             $this->db->where_in('shop_cats.cat_id' , $cat);
         }

但我收到语法错误!

Answer 1

尝试

$this->db->distinct('shops.*');
$this->db->from('shops');
$this->db->join('shop_cats', 'shop_cats.shop_id = shops.id', 'left');
$this->db->where('shop_cats.cat_id', $cat);
$this->db->limit($limit , $offset);
$res = $this->db->get();