Question

我想在where where条件下传递多个id怎么做？

查询从数据库中提取org_id。

现在我想在我的情况下通过那个，那该怎么做？

我试过foreach循环：

以下是代码：

$devices =  $this->activation_m->get_activated_devices();


    $org_id = array(); // create org_id array
        foreach ($devices as $row)
        {

         $org_id[]= $row['org_id']; // assign ids into array

            //$companys =$this->data['devices'] = $this->activation_m->get_company($org_id); // don't call again and again
        }

     //Now $org_id is array


    $companys =$this->data['devices'] = $this->activation_m->get_company($org_id);  
    echo $this->db->last_query();

型号代码

    public function get_activated_devices()
    {
        $this->db->select('*');
        $this->db->join('sitekey','sitekey.site_key = activation.site_key');
        $this->db->from('activation');
        $query =$this->db->get();
        $result = $query->result_array(); 
        return $result; 




    }

    public function get_company($org_id)
    {

        $this->db->select('*');
        $this->db->join('sitekey','sitekey.org_id = company.id');
        $this->db->join('activation','sitekey.site_key = activation.site_key');
        $this->db->where('company.id IN',(implode(',',$org_id))); // see the change here
        $this->db->from('company');
        $query =$this->db->get();
        $result = $query->result_array(); 
        return $result; 
    }

现在我的查询目前只传递了一个org_id，我想传递我从第一个查询中获得的所有org_id。

Answer 1

您可以使用来自活动记录codeigniter的where_in

as

$devices =  $this->activation_m->get_activated_devices();
$org_id = array();  
foreach ($devices as $row)
{    
 $org_id[] = $row['org_id'];           
}
 $companys =$this->data['devices'] = $this->activation_m->get_company($org_id);
if( $companys->num_rows() > 0 )
{
   echo "<pre>";
   print_r( $companys->result());
   echo "</pre>";
}

对于模型

public function get_company($org_ids = array())
{
    $this->db->select('*');
    $this->db->join('sitekey','sitekey.org_id = company.id');
    $this->db->join('activation','sitekey.site_key = activation.site_key');
    $this->db->where_in('company.id', $org_ids );  //this is condition       
    $this->db->from('company');
    return $this->db->get();
}

Answer 2

您可以将codeigniter的$this->db->or_where()用于此目的。只需遍历组织ID的数组并应用or_where条件。

    $this->db->select('*');
    $this->db->join('sitekey','sitekey.org_id = company.id');
    $this->db->join('activation','sitekey.site_key = activation.site_key');
    foreach($org_id as $org)
    {    // where $org is the instance of one object of active record
         $this->db->or_where('company.id',$org);
    }
    $this->db->from('company');
    $query =$this->db->get();
    $result = $query->result_array(); 
    return $result;

另一种方法是通过像这样遍历数组并在字符串中附加where子句来创建自定义查询。

如何在条件codeigniter中传递多个Id数组？

2 个答案: