每15分钟一次GROUP BY时间戳，包括缺失条目 [英] GROUP BY timestamp every 15 minutes including missing entries

问题描述

我想每隔15分钟将名为'reports'的给定表格中的所有报告进行分组，包括没有填充报告的时间间隔。示例：

I want to group all reports from a given table named 'reports' by every 15 minutes, including intervals where no reports where filled. Example:

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

选择查询后，我应该有：

After the select query I should have:

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：

What I tried is the following query but this doesn't include the missing 15-minute-intervals:

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

推荐答案

如果你真的想在mysql中这样做，这应该在一定程度上 - 取决于你想如何组织15分钟的时间间隔。我已经从|输出了的时间时间| |报告数量，因为我认为它没有意义，甚至有点欺骗性，不能输出特定的时间戳，并将时间范围的计数与它关联在一起。

If you really want to do it in mysql, this should work to some extent - depending on how you want to organise your 15 minute intervals. I've given the output as time from | time to | number of reports because i think its meaningless and even a little deceptive to output a specific timestamp and associate a count of a time range with it.

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;

这篇关于每15分钟一次GROUP BY时间戳，包括缺失条目的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！