php - 如何在where条件codeigniter中传递多个Id数组？

Question

我想在where条件下传递多个id怎么办？

查询从数据库中获取 org_id。

现在我想在我的 where 条件下传递它，那么该怎么做呢？

我尝试了 foreach 循环:

下面是我的代码:

$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();

模型代码

    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; 
    }

目前我的查询只传递一个 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 active-records codeigniter 的 where_in

作为

$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>";
}

和模型

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();
}