Question

这是我的数据，每小时可能有几个Id

UKey - Primary Key
Epoch - epoch
Id - the transaction ID

查找数据库中所有日期的每小时的第一个交易ID 。

数据：

UKey   Epoch                Id
1     11:00:23 21-01-2016 xxx
2     11:01:33 21-01-2016 yyy
3     11:33:22 21-01-2016 zzz
4     12:03:41 21-01-2016 aaa
5     12:23:32 21-01-2016 bbb
6     12:34:43 21-01-2016 ccc
7     13:03:21 21-01-2016 ddd
8     13:04:22 21-01-2016 eee
9     14:00:01 21-01-2016 fff
10    16:00:01 23-01-2016 ggg

寻找这样的结果：

UKey   Epoch                Id
1     11:00:23 21-01-2016 xxx
4     12:03:41 21-01-2016 aaa
7     13:03:21 21-01-2016 ddd
9     14:00:01 21-01-2016 fff
10    16:00:01 23-01-2016 ggg

感谢您对此的投入，请确保此处的数据是一个示例，可能会有很多天和月，因此日期可能会有所不同

Answer 1

DROP TABLE IF EXISTS my_table;

CREATE TABLE my_table
(ukey INT NOT NULL AUTO_INCREMENT PRIMARY KEY
,epoch DATETIME NOT NULL
,id CHAR(12)
);

INSERT INTO my_table (epoch,id) VALUES
('2016-01-21 11:00:23', 'xxx'),
('2016-01-21 11:01:33', 'yyy'),
('2016-01-21 11:33:22', 'zzz'),
('2016-01-21 12:03:41', 'aaa'),
('2016-01-21 12:23:32', 'bbb'),
('2016-01-21 12:34:43', 'ccc'),
('2016-01-21 13:03:21', 'ddd'),
('2016-01-21 13:04:22', 'eee'),
('2016-01-21 14:00:01', 'fff'),
('2016-01-23 16:00:01', 'ggg');

SELECT x.* 
  FROM my_table x 
  JOIN 
     ( SELECT DATE_FORMAT(epoch,'%Y-%M-%D %h')dt,MIN(ukey) min_ukey 
         FROM my_table 
        GROUP BY dt
     ) y 
    ON y.dt = DATE_FORMAT(x.epoch,'%Y-%M-%D %h') 
   AND y.min_ukey = x.ukey;

+------+---------------------+------+
| ukey | epoch               | id   |
+------+---------------------+------+
|    7 | 2016-01-21 13:03:21 | ddd  |
|    9 | 2016-01-21 14:00:01 | fff  |
|    1 | 2016-01-21 11:00:23 | xxx  |
|    4 | 2016-01-21 12:03:41 | aaa  |
|   10 | 2016-01-23 16:00:01 | ggg  |
+------+---------------------+------+

选择并获得每小时的结果

1 个答案: