どのように私は毎時の最初の行を得ることができますか？

Question

MySQL []>列を読み込みから表示します。

| Field  | Type  | Null | Key | Default | Extra   | 
| cont  | int(4)  | NO | PRI | NULL | auto_increment | 
| id   | int(4)  | YES | MUL | NULL |    | 
| sensor_num | int(4)  | YES |  | NULL |    | 
| value  | float(5,2) | YES |  | NULL |    | 
| time  | datetime | YES |  | NULL |    | 

select * from reading;

18570 | 1 |   1 | 23.93 | 2017-08-14 12:45:01 | 
| 18571 | 1 |   1 | 23.93 | 2017-08-14 12:46:01 | 
| 18572 | 1 |   1 | 23.93 | 2017-08-14 12:47:01 | 
| 18573 | 1 |   1 | 23.93 | 2017-08-14 12:48:01 | 
| 18574 | 1 |   1 | 23.93 | 2017-08-14 12:49:02 | 
| 18575 | 1 |   1 | 23.93 | 2017-08-14 12:50:02 | 
| 18576 | 1 |   1 | 23.93 | 2017-08-14 12:51:02 | 
| 18577 | 1 |   1 | 23.93 | 2017-08-14 12:52:02 | 
| 18578 | 1 |   1 | 23.93 | 2017-08-14 12:53:02 | 
| 18579 | 1 |   1 | 23.93 | 2017-08-14 12:54:02 | 
| 18580 | 1 |   1 | 23.93 | 2017-08-14 12:55:02 | 
| 18581 | 1 |   1 | 23.93 | 2017-08-14 12:56:02 | 
| 18582 | 1 |   1 | 23.93 | 2017-08-14 12:57:02 | 

| 18689 | 1 |   1 | 25.88 | 2017-08-14 14:54:56 | 
| 18690 | 1 |   1 | 25.88 | 2017-08-14 14:55:56 | 
| 18691 | 1 |   1 | 25.88 | 2017-08-14 14:56:56 | 
| 18692 | 1 |   1 | 25.88 | 2017-08-14 14:57:56 | 
| 18693 | 1 |   1 | 25.88 | 2017-08-14 14:58:56 | 
| 18694 | 1 |   1 | 25.88 | 2017-08-14 14:59:56 | 
| 18695 | 1 |   1 | 25.88 | 2017-08-14 15:00:56 | 
| 18696 | 1 |   1 | 25.88 | 2017-08-14 15:01:56 | 
| 18697 | 1 |   1 | 25.88 | 2017-08-14 15:02:56 | 
| 18698 | 1 |   1 | 25.88 | 2017-08-14 15:03:56 | 
| 18699 | 1 |   1 | 25.88 | 2017-08-14 15:04:56 | 
| 18700 | 1 |   1 | 25.88 | 2017-08-14 15:05:56 | 

Answer 1

あなたは、各時間の最初の行をしたい場合は、あなたが時間によって分（時間）グループと一致した行を選択することができます。

select * 
    from reading a 
    inner join (
    select date(time) t_date, min(time) t_time 
    from reading 
    group by date(time), hour(time) 
) t on t.t_time = a.time