Question

两张桌子：第一个 - kons，二阶。由字段kons.id和order.id_kons链接这是我的问题：

$this->db->select("k.id as k_id, DATE_FORMAT(k.datecreate, '%d/%m/%Y %H:%i:%s') as k_dt, o.id as o_id",false);
 $this->db->join('orders as o', 'o.id_kons = k.id','inner');
 $this->db->from('kons as k');<br>
 return $this->db->get()->result_array();

结果是（例如，剪切）：

array(20) {
  [0]=>
  array(3) {
    ["k_id"]=>
    string(2) "45"
    ["k_dt"]=>
    string(19) "22/02/2014 15:41:35"
    ["o_id"]=>
    string(3) "533"
  }

  } .....

但我需要这样的东西：

array(x) {
  [0]=>
  array(3) {
    ["k_id"]=>
    string(2) "45"
    ["k_dt"]=>
    string(19) "22/02/2014 15:41:35"
    ["o_id"]=>
    array(3) "533,534,536" or string(x) "533,534,536"
  }

Answer 1

您可以使用GROUP_CONCAT(expr)来获取逗号分隔的顺序ID，因为默认情况下字符长度限制设置为1024但可以增加

$this->db->select("k.id as k_id, 
DATE_FORMAT(k.datecreate, '%d/%m/%Y %H:%i:%s') as k_dt, 
GROUP_CONCAT(o.id) as o_id",false);
$this->db->join('orders as o', 'o.id_kons = k.id','inner');
$this->db->from('kons as k');
$this->db->group_by("k.id");
return $this->db->get()->result_array();

Codeigniter，Active记录，将第二个表值连接为数组

1 个答案: