mysqli查詢[活動記錄Codeigniter]

Question

我有一個時間表如下

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

日期為Y|m|d格式，時間為24 hrs格式， status= 0表示未完成

讓current date 2015|9|19 and time 10:05:3現在，我希望獲取當前日期和時間用戶未完成的所有任務，以便結果看起來像下面這樣，總未完成任務為好

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

請建議同時使用mysqli和CodeIgniter's Active-records查詢。
問候.....

Answer 1

您應該嘗試使用此方法進行活動記錄：

$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();

對於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>";

        }