Question

我跟着表。

       apartments
       id name slug created modified

       apartment_amenities
       id name slug apartment_id created modified

       apartment_activities
       id name slug apartment_id created modified

在视图中我想要这样的东西。

       no apartment_name    amenities       activities

       1  shobha_comnplex   party hall      pamplets
                            swimming pool   banners
                            play area       boards

       2  navami_comnplex   party hall      boards
                            swimming pool   banners
                            club house      pamplets

在模特中我试过这样的事。

        $this->db->select('apartments.id, apartments.slug, apartments.name, apartment_amenities.name as amenity_name, apartment_activities.name as activity_name');
        $this->db->from($this->_table);
        $this->db->join('apartment_amenities', 'apartment_amenities.apartment_id = apartments.id', 'left');
        $this->db->join('apartment_activities', 'apartment_activities.apartment_id = apartments.id', 'left'); 
        return $this->db->get();

但即使公寓有许多设施和活动，我只能获得单一的设施和活动。结果如下。

Array
(
    [0] => 
        (
            [id] => 1
            [slug] => shobha_complex
            [name] => shobha complex
            [amenity_name] => party hall
            [activity_name] => pamplets
        ),

    [1] => 
        (
            [id] => 1
            [slug] => navami_complex
            [name] => navami complex
            [amenity_name] => party hall
            [activity_name] => boards
        )
)

我希望结果如下所示。

    Array
    (
        [0] => 
            (
                [id] => 1
                [slug] => shobha_complex
                [name] => shobha complex
                [amenities] => Array(
                    [0] => 
                         (
                            [name] => party hall
                         ),
                    [1] => 
                         (
                            [name] => swimming pool
                         ),
                    [2] => 
                         (
                            [name] => play area
                         )
                ),
                [activities] => Array(
                    [0] => 
                         (
                            [name] => pamplets
                         ),
                    [1] => 
                         (
                            [name] => banners
                         ),
                    [2] => 
                         (
                            [name] => boards
                         )
                )
            ),

        [1] => 
            (
                [id] => 1
                [slug] => navami_complex
                [name] => Navami complex
                [amenities] => Array(
                    [0] => 
                         (
                            [name] => party hall
                         ),
                    [1] => 
                         (
                            [name] => swimming pool
                         ),
                    [2] => 
                         (
                            [name] => club house
                         )
                ),
                [activities] => Array(
                    [0] => 
                         (
                            [name] => boards
                         ),
                    [1] => 
                         (
                            [name] => banners
                         ),
                    [2] => 
                         (
                            [name] => pamplets
                         )
                )
            ),
    )

请建议我如何获得解决方案。这项工作将更受赞赏。

Answer 1

您可以在select上使用group concat并将其用作分隔符|因此，您将只获得两行，并在视图中使用分隔符

拆分列

$this->db->select("GROUP_CONCAT(partment_amenities.name SEPARATOR '|') as amenity_name, ...... ", FALSE);

第二个参数FALSE表示不保护标识符有关组concate的更多信息，请参阅链接： https://dev.mysql.com/doc/refman/5.0/en/group-by-functions.html#function_group-concat

Answer 2

删除$this->db->group_by方法调用。这只会为每个公寓返回一个结果。如果你删除它，你应该得到你想要的所有设施等。

唯一的问题是你得到这样的结果

   no apartment_name    amenities       activities

   1  shobha_comnplex   party hall      pamplets
   1  shobha_comnplex   swimming pool   banners
   1  shobha_comnplex   play area       boards

   2  navami_comnplex   party hall      boards
   2  navami_comnplex   swimming pool   banners
   2  navami_comnplex   club house      pamplets

Answer 3

要设置一个合适的数组，我们会以你想要的格式建议你有一些模型。

让我们设置一个场景

<强>控制器：

$apartments = $this->apartment_model->getApartments(); // Get All Apartments
foreach($apartments as &$apartment)
{
  $apartment->amenities = $this->apartment_model->getAmenities($apartment->id);
  $apartment->activities= $this->apartment_model->getActivities($apartment->id);

  // Add / Modify any addition properties
}

// Do something with the $apartments array

<强>型号：

class Apartment_model extends model
{
  function getApartments()
  {
    $this->db->select('id, slug, name');
    $this->db->from('apartments');
    $query = $this->db->get();
    return $query->result();
  }
  function getAmenities($apartment_id)
  {
    $this->db->select('name');
    $this->db->from('apartment_amenities');
    $this->db->where('id', $apartment_id);
    $query = $this->db->get();
    return $query->result();
  }
  function getActivities($apartment_id)
  {
    $this->db->select('name');
    $this->db->from('apartment_activities');
    $this->db->where('id', $apartment_id);
    $query = $this->db->get();
    return $query->result();
  }
}

当然要记住，这还没有加载视图或者将数据移动到控制器的任何地方，但它至少会设置数组。

我希望这有助于您入门。

用于连接表的CodeIgniter活动记录

3 个答案: