Question

我想知道每小时的使用量是基于以下数据：

TIME        DATA
00:00:24    3198
00:17:42    3199
00:44:54    3200
01:24:24    3201
01:52:24    3202
02:27:18    3203
02:51:06    3204
03:28:30    3205

当我使用如下查询时：

SELECT
    left(TIME,2) AS hour,
    max(DATA) - min(DATA) as usage
    FROM my_table
GROUP BY hour

我得到的值不正确，因为MySQL只查看小时范围中与最大值相比的最小值。结果是：

hour     usage
00       2
01       1
02       1
03       0

如果我手动查看数据并手动求和，我希望：

TIME        DATA    Usage (summed by hand)
00:00:24    3198        
00:17:42    3199        
00:44:54    3200    2 
01:24:24    3201        
01:52:24    3202    3  
02:27:18    3203        
02:51:06    3204    2
03:28:30    3205    1

有没有办法直接在MySQL中解决这个问题，还是需要在PHP中创建一个循环？（编辑查询和澄清结果及预期结果）

Answer 1

试试这个：

   select left(time, 2), count(*) from <your table name> group by left(time, 2)

Answer 2

考虑这个解决方案：

SELECT LEFT(time, 2) AS `hour`, COUNT(LEFT(time, 2)) AS `count`
FROM tbl
GROUP BY `hour`

您必须将tbl替换为您的表格名称。

Answer 3

检查以下查询：

SELECT A.TIME , A.DATA , CASE WHEN B.USAGE IS NULL THEN NULL ELSE USAGE END AS USAGE
FROM
USAGE AS A
LEFT OUTER JOIN
(
SELECT LEFT(TIME , 2) AS TIMEINHOUR ,  
MAX(TIME) AS MAXTIME , MAX(DATA) - MIN(DATA) AS USAGE
FROM USAGE
GROUP BY LEFT(TIME , 2)
) AS B ON A.TIME = B.MAXTIME

MySQL基于运行总计的每小时使用量

3 个答案: