mysqli查询[活动记录Codeigniter]

Question

我有一个时间表如下

id task to-do-date to-do-time status 1 X 2015|9|19 21:10:1 0 2 Y 2015|9|19 09:05:3 0 3 Z 2015|9|17 08:12:3 0 4 A 2015|9|16 23:10:5 0

日期为Y|m|d格式，时间为24 hrs格式， status= 0表示未完成

让current date 2015|9|19 and time 10:05:3现在，我希望获取当前日期和时间用户未完成的所有任务，以便结果看起来像下面这样，总未完成任务为好

id task to-do-date to-do-time status 2 Y 2015|9|19 09:05:3 0 3 Z 2015|9|17 08:12:3 0 4 A 2015|9|16 23:10:5 0 total Not-Done-taks =3

请建议同时使用mysqli和CodeIgniter's Active-records查询。
问候.....

Answer 1

您应该尝试使用此方法进行活动记录：

$this->db->select();
$this->db->from('your_table_name');
$this->db->where('to-do-date <= ',date('Y-m-d'));
$this->db->where('to-do-time <= ',date('H:i:s'));
$this->db->where('status',0);
return $this->db->get()->result();

对于mysqli：

$sql = "SELECT * FROM your_table_name WHERE to-do-date <= '".date('Y-m-d')."' AND to-do-time <= '".date('H:i:s')."' AND status = 0";

Answer 2

$this->db->select();
        $this->db->from('your_table_name');
        $this->db->where('to_do_date < ',date('Y-m-d'));
        $this->db->where('status','0');
        $this->db->or_where('to_do_date',date('Y-m-d'));
        $this->db->where('to_do_time < ',date('H:i:s'));
        $this->db->where('status',0);           
        $skipped = $this->db->get()->result_array();
        foreach ($skipped as $skp) {
            echo $skp['task'], $skp['to_do_date'],$skp['to_do_time'],$skp['status'], "</br>";

        }