Question

情况（简化）：

TABLEA

id |   date               | val
------------------------------
0    2018-02-19 00:01:00  | 10
1    2018-02-19 00:02:00  | 10
2    2018-02-19 00:03:00  | 10
..   2018-02-19 23:59:00  | 10

我需要执行一个查询，每小时返回列val的SUM。

这是查询

SELECT `AllHours`.`hour` , COALESCE(SUM(`A`.`val`),0) AS `A`.`total`
FROM `tableA` AS `A`

RIGHT JOIN (
    SELECT  0 AS `hour`
    UNION ALL SELECT  1 UNION ALL SELECT  2 UNION ALL SELECT  3
         UNION ALL SELECT  4 UNION ALL SELECT  5 UNION ALL SELECT  6
         UNION ALL SELECT  7 UNION ALL SELECT  8 UNION ALL SELECT  9
         UNION ALL SELECT 10 UNION ALL SELECT 11 UNION ALL SELECT 12
         UNION ALL SELECT 13 UNION ALL SELECT 14 UNION ALL SELECT 15
         UNION ALL SELECT 16 UNION ALL SELECT 17 UNION ALL SELECT 18
         UNION ALL SELECT 19 UNION ALL SELECT 20 UNION ALL SELECT 21
         UNION ALL SELECT 22 UNION ALL SELECT 23
  )      AS `AllHours` ON HOUR(`A`.`date`) = `AllHours`.`hour`


WHERE `A`.`date` BETWEEN '2018-02-19 00:00:00' AND '2018-02-19 23:59:59' OR `A`.`date` IS NULL

GROUP BY `AllHours`.`hour`
ORDER BY `AllHours`.`hour`

结果

查询有效，但缺少第11小时。

重要提示 如果没有数据，我需要所有的时间。否则我就不会加入RIGHT JOIN，这就足够了GROUP BY HOUR（约会）。

Answer 1

考虑使用数字表 - 请参阅下面的链接。这将为您提供所需的输出：

TESTDATA：

CREATE TABLE T   (`id` int, `dt` datetime, `val` int);        
INSERT INTO T    (`id`, `dt`, `val`) 
VALUES
    (0, '2018-02-19 00:01:00', 10),
    (1, '2018-02-19 00:02:00', 10),
    (2, '2018-02-19 00:03:00', 10),
    (4, '2018-02-19 01:01:00', 10),
    (5, '2018-02-19 01:02:00', 10),
    (6, '2018-02-19 02:03:00', 10)
;

SQL：

select 
  lpad(cast(HH as char(2)),2,'0') as hour,  
  sum(val) as sumVal from
(
    select 
      EXTRACT(HOUR from dt) AS HH,
      val
    from T 
    WHERE dt >= '2018-02-19 0:0:0' and dt < '2018-02-20 0:0:0'

    UNION ALL SELECT  0,0
    UNION ALL SELECT  1,0 UNION ALL SELECT  2,0  UNION ALL SELECT  3,0
    UNION ALL SELECT  4,0 UNION ALL SELECT  5,0  UNION ALL SELECT  6,0
    UNION ALL SELECT  7,0 UNION ALL SELECT  8,0 UNION ALL SELECT  9,0
    UNION ALL SELECT 10,0 UNION ALL SELECT 11,0 UNION ALL SELECT 12,0
    UNION ALL SELECT 13,0 UNION ALL SELECT 14,0 UNION ALL SELECT 15,0
    UNION ALL SELECT 16,0 UNION ALL SELECT 17,0 UNION ALL SELECT 18,0
    UNION ALL SELECT 19,0 UNION ALL SELECT 20,0 UNION ALL SELECT 21,0
    UNION ALL SELECT 22,0 UNION ALL SELECT 23,0
) as m
group by lpad(cast(HH as char(2)),2,'0')

获得输出：

hour    sumVal
00  30
01  20
02  10
03  0
04  0
05  0
06  0
07  0
08  0
09  0
10  0
11  0
12  0
13  0
14  0
15  0
16  0
17  0
18  0
19  0
20  0
21  0
22  0
23  0

如果您更频繁地需要这个，请创建一个数字表并使用您已有的连接语法。没有数字表，你可以在没有加入的情况下结合所有。

SO-Readups：

使用数字表，上面的大联盟可以重写为

UNION ALL SELECT  num, 0 from numbers where num between 0 and 23

或者您可以使用连接和您的合并语法。

Answer 2

试一试：

SELECT
`AllHours`.`tmp_hour`,
COALESCE(SUM(`val`),0) AS `total`
    FROM
      tableA
    RIGHT JOIN (
    SELECT  0 AS `tmp_hour`
    UNION ALL SELECT  1 UNION ALL SELECT  2 UNION ALL SELECT  3
         UNION ALL SELECT  4 UNION ALL SELECT  5 UNION ALL SELECT  6
         UNION ALL SELECT  7 UNION ALL SELECT  8 UNION ALL SELECT  9
         UNION ALL SELECT 10 UNION ALL SELECT 11 UNION ALL SELECT 12
         UNION ALL SELECT 13 UNION ALL SELECT 14 UNION ALL SELECT 15
         UNION ALL SELECT 16 UNION ALL SELECT 17 UNION ALL SELECT 18
         UNION ALL SELECT 19 UNION ALL SELECT 20 UNION ALL SELECT 21
         UNION ALL SELECT 22 UNION ALL SELECT 23
  )      AS `AllHours` ON DATE_FORMAT(dt, '%H') = `AllHours`.`tmp_hour`

    WHERE  `dt` BETWEEN '2018-02-19 00:00:00' AND '2018-02-19 23:59:59' OR `dt` IS NULL

    GROUP BY `AllHours`.`tmp_hour`
    ORDER BY `AllHours`.`tmp_hour`

Mysql，SUM列和GROUP BY HOUR

2 个答案: