我有一个时间表,如下所示
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
其中日期为
Y|m|d
格式且时间为24 hrs
格式,status= 0
表示未完成
让我们current date 2015|9|19 and time 10:05:3
现在,我希望获取当前日期和时间没有完成的所有任务,因此结果可能看起来像完全没有完成的任务一样
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
mysqli
和CodeIgniter's Active-records
查询。
答案 0 :(得分:1)
您应该尝试使用此活动记录:
$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();
和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";
答案 1 :(得分:0)
$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>";
}