Question

我正在尝试选择ID并使用此

将其传递给更新

$query = $this->db->query("SELECT GROUP_CONCAT(a.sponsor_id) as sponstr FROM (select sponsor_id from sponsor  WHERE (pay_success = 'yes')AND (end_date_time > NOW()) and ((country_id = 1 and state_id = 24) or city_id = 123)
order by rand() limit 0,10) a");

    if($query->num_rows()>0)
    {
     foreach($query->result() as $sponsorids)
     {

        $data['se_count'] = 0;
        $this->db->where_in('sponsor_id',$sponsorids->sponstr);
        $this->db->update('sponsor',$data);

     }
     }

但所有ID都不会更新，只有第一个ID会更新。

where_in生成下面的代码

 WHERE sponsor_id IN ('5,4,2,3,1')

我认为它应该是

  WHERE sponsor_id IN (5,4,2,3,1)

我在这里遗漏了什么，或者我做错了什么，显然我知道我是。请帮忙

Answer 1

你应该在那里传递数组。所以不要通过$sponsorids->sponstr而是explode(',', $sponsorids->sponstr)

此外，这似乎是一个糟糕的数据库设计决策，花一些时间看看多对多的概念

将select选择传入where_in Codeigniter

1 个答案: