这是按小时分组的公式吗?

发布于 2024-12-15 15:07:35 字数 294 浏览 4 评论 0原文

N - 要分组的小时数

DATETIME列值(例如,2011-10-08 21:23:43)

注意:对于Mysql DB

GROUP BY date( `DATETIME` ) , N* floor( date_format( `DATETIME` , '%H' ) /N ))
  1. 是这样的按小时分组公式?

  2. 有按日/月分组的公式吗?

N - the number of hours to be grouped

DATETIME column value (eg, 2011-10-08 21:23:43)

Note: for Mysql DB

GROUP BY date( `DATETIME` ) , N* floor( date_format( `DATETIME` , '%H' ) /N ))
  1. Is this Group by hours formula?

  2. Any formula for group by day/month?

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維他命╮ 2024-12-22 15:07:35
SELECT MONTH(dt) as m, DAY(dt) as d, COUNT(*) as cnt
FROM (
SELECT '2011-10-11 10:00:00' as dt
UNION ALL
SELECT '2011-10-11 20:00:00' as dt
UNION ALL
SELECT '2011-10-12 10:00:00' as dt
UNION ALL
SELECT '2011-10-12 20:00:00' as dt
UNION ALL
SELECT '2011-10-13 10:00:00' as dt
UNION ALL
SELECT '2011-10-13 20:00:00' as dt
UNION ALL
SELECT '2011-11-13 10:00:00' as dt
UNION ALL
SELECT '2011-11-13 20:00:00' as dt
) as dates
GROUP BY m, d;

结果是:

+------+------+-----+
| m    | d    | cnt |
+------+------+-----+
|   10 |   11 |   2 |
|   10 |   12 |   2 |
|   10 |   13 |   2 |
|   11 |   13 |   2 |
+------+------+-----+
SELECT MONTH(dt) as m, DAY(dt) as d, COUNT(*) as cnt
FROM (
SELECT '2011-10-11 10:00:00' as dt
UNION ALL
SELECT '2011-10-11 20:00:00' as dt
UNION ALL
SELECT '2011-10-12 10:00:00' as dt
UNION ALL
SELECT '2011-10-12 20:00:00' as dt
UNION ALL
SELECT '2011-10-13 10:00:00' as dt
UNION ALL
SELECT '2011-10-13 20:00:00' as dt
UNION ALL
SELECT '2011-11-13 10:00:00' as dt
UNION ALL
SELECT '2011-11-13 20:00:00' as dt
) as dates
GROUP BY m, d;

results in:

+------+------+-----+
| m    | d    | cnt |
+------+------+-----+
|   10 |   11 |   2 |
|   10 |   12 |   2 |
|   10 |   13 |   2 |
|   11 |   13 |   2 |
+------+------+-----+
~没有更多了~
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