每 15 分钟 GROUP BY 时间戳，包括丢失的条目答案

【问题标题】：GROUP BY timestamp every 15 minutes including missing entries每 15 分钟 GROUP BY 时间戳，包括丢失的条目
【发布时间】：2015-04-18 07:29:50
【问题描述】：

我想每 15 分钟对名为“报告”的给定表格中的所有报告进行分组，包括没有填写报告的时间间隔。示例：

id      timestamp 
---------------------------
1        2015-04-16 20:52:04
2        2015-04-16 20:53:04
3        2015-04-16 20:54:04
4        2015-04-16 19:52:04
5        2015-04-17 22:24:56
6        2015-04-17 22:27:09
7        2015-04-18 06:48:41

在选择查询之后我应该有：

timestamp       count
----------------------
20:52:04         3
21:07:04         0
21:22:04         0
21:37:04         0
21:52:04         0
22:07:04         0
22:22:04         2
22:37:04         0
22:52:04         0
......
06:52:04         1
07:07:04         0

我尝试的是以下查询，但这不包括缺少的 15 分钟间隔：

select created_at , count(id) AS count 
from `reports` 
where `company_id` = '3' 
group by UNIX_TIMESTAMP(created_at) DIV 900 
order by `created_at` asc

【问题讨论】：

最简单的方法是创建一个包含 15 分钟间隔时间戳的表并加入它
我想你想要从过去 15 分钟开始的计数，而不仅仅是第 15 分钟？
我认为您将数据存储和检索与数据显示混淆了

标签： mysql select group-by timestamp

【解决方案1】：

如果您真的想在 mysql 中执行此操作，这应该在一定程度上起作用 - 取决于您希望如何组织 15 分钟间隔。我将输出设为time from | time to | number of reports，因为我认为输出特定时间戳并将时间范围的计数与其关联起来毫无意义，甚至有点欺骗性。

SELECT CONCAT (
        lpad(hours.a, 2, "0"),
        ":",
        lpad(minutes.a * 15, 2, "0"),
        ":00"
        ) AS 'from',
    CONCAT (
        lpad(hours.a, 2, "0"),
        ":",
        lpad((minutes.a * 15) + 14, 2, "0"),
        ":59"
        ) AS 'to',
    count(r.id)
FROM (
    SELECT 0 AS a

    UNION

    SELECT 1 AS a

    UNION

    SELECT 2 AS a

    UNION

    SELECT 3 AS a

    UNION

    SELECT 4 AS a

    UNION

    SELECT 5 AS a

    UNION

    SELECT 6 AS a

    UNION

    SELECT 7 AS a

    UNION

    SELECT 8 AS a

    UNION

    SELECT 9 AS a

    UNION

    SELECT 10 AS a

    UNION

    SELECT 11 AS a

    UNION

    SELECT 12 AS a

    UNION

    SELECT 13 AS a

    UNION

    SELECT 14 AS a

    UNION

    SELECT 15 AS a

    UNION

    SELECT 16 AS a

    UNION

    SELECT 17 AS a

    UNION

    SELECT 18 AS a

    UNION

    SELECT 19 AS a

    UNION

    SELECT 20 AS a

    UNION

    SELECT 21 AS a

    UNION

    SELECT 22 AS a

    UNION

    SELECT 23 AS a
    ) hours
INNER JOIN (
    SELECT 0 AS a

    UNION

    SELECT 1 AS a

    UNION

    SELECT 2 AS a

    UNION

    SELECT 3 AS a
    ) minutes
LEFT JOIN reports r ON hours.a = hour(r.t)
    AND minutes.a = floor(minute(r.t) / 15)
GROUP BY hours.a,
    minutes.a;

【讨论】：