mySQL查询以返回给定时期的最短状态持续时间

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英文:

mySQL query to return minimum state duration for a given period

问题

我有一个MySQL数据库表中的以下数据,我想要计算给定时期内 state = 1 的最短持续时间,以放入Grafana的统计数据中。

在这个表片段中,期望的结果是第10行和第8行之间的timestampdiff(),即2分钟。

id loc temp state datetime
1 room_1 19.4 0 2022-12-29 18:14:09
2 room_1 19.3 0 2022-12-29 18:15:09
3 room_1 19.2 1 2022-12-29 18:16:09
4 room_1 19.2 1 2022-12-29 18:17:09
5 room_1 19.2 1 2022-12-29 18:18:09
6 room_1 19.2 0 2022-12-29 18:19:10
7 room_1 19.2 0 2022-12-29 18:20:10
8 room_1 19.1 1 2022-12-29 18:21:10
9 room_1 19.0 1 2022-12-29 18:22:10
10 room_1 19.0 0 2022-12-29 18:23:10

我已经尝试了许多有问题的解决方案,但出于明显的原因,我不会在这里发布。

关于时期部分,可以使用Grafana的全局变量WHERE $__timeFilter(datetime),并使用它可以计算 持续时间 WHERE state = 1 AND $__timeFilter(datetime),因为记录生成是基于1分钟的基础,这使得对二进制0/1状态列求和变得容易(显然不需要state = 1)。

我完全迷失在尝试在时期内处理持续时间。

英文:

I have the following data in a MySQL database table, and I'd like to calculate the minimum duration where state = 1 for a given period, to place into a Grafana stat.

Within this table snippet, the desired result would be the timestampdiff() between row 10 and 8, i.e. 2 minutes.

id loc temp state datetime
1 room_1 19.4 0 2022-12-29 18:14:09
2 room_1 19.3 0 2022-12-29 18:15:09
3 room_1 19.2 1 2022-12-29 18:16:09
4 room_1 19.2 1 2022-12-29 18:17:09
5 room_1 19.2 1 2022-12-29 18:18:09
6 room_1 19.2 0 2022-12-29 18:19:10
7 room_1 19.2 0 2022-12-29 18:20:10
8 room_1 19.1 1 2022-12-29 18:21:10
9 room_1 19.0 1 2022-12-29 18:22:10
10 room_1 19.0 0 2022-12-29 18:23:10

I've tried numerous faulty solutions from chatGPT that I won't post here for obvious reasons.

The period part is simple enough using the Grafana global variable WHERE $__timeFilter(datetime) and using that I can calculate the total duration WHERE state = 1 AND $__timeFilter(datetime) because the record generation is on a 1-minute basis which makes summing the binary 0/1 state column an easy out (obviously doesn't require the state = 1).

I'm completely lost trying to work with duration periods within a period.

答案1

得分: 0

以下是您需要的查询要求的翻译:

WITH temp_prev_state AS (
SELECT id, loc, temp, state, datetime_, LAG(state,1)  OVER(  ORDER BY datetime_  ) AS prev_state FROM Temperature
), latest_on_off_stateSession AS (
(SELECT * from temp_prev_state where state = '0' and prev_state = '1' ORDER BY datetime_ DESC LIMIT 1 ) 
UNION
(SELECT * from temp_prev_state where state = '1' and prev_state = '0' ORDER BY datetime_ DESC LIMIT 1 )
), lat_on_off_prev_dt AS (
SELECT *, LAG(datetime_,1)  OVER(  ORDER BY datetime_  ) AS prev_datetime_ FROM latest_on_off_stateSession
)
select id, loc, temp, state, datetime_, ABS(TIMESTAMPDIFF(MINUTE,datetime_ ,prev_datetime_)) AS abs_Time_Diff_Mins FROM lat_on_off_prev_dt

查询的输出结果:
mySQL查询以返回给定时期的最短状态持续时间

在此处查看演示:https://dbfiddle.uk/Sm7q9P7Y

英文:

The query requirements that you had was bit complex, I had to use windows function - LAG to get the results. Below is the query that you need.

WITH temp_prev_state AS (
SELECT id, loc, temp, state, datetime_, LAG(state,1)  OVER(  ORDER BY datetime_  ) AS prev_state FROM Temperature
), latest_on_off_stateSession AS (
(SELECT * from temp_prev_state where state = '0' and prev_state = '1' ORDER BY datetime_ DESC LIMIT 1 ) 
UNION
(SELECT * from temp_prev_state where state = '1' and prev_state = '0' ORDER BY datetime_ DESC LIMIT 1 )
), lat_on_off_prev_dt AS (
SELECT *, LAG(datetime_,1)  OVER(  ORDER BY datetime_  ) AS prev_datetime_ FROM latest_on_off_stateSession
)
select id, loc, temp, state, datetime_, ABS(TIMESTAMPDIFF(MINUTE,datetime_ ,prev_datetime_)) AS abs_Time_Diff_Mins FROM lat_on_off_prev_dt


The output of the query:
mySQL查询以返回给定时期的最短状态持续时间

Check the demo here

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  • 本文由 发表于 2023年3月12日 08:58:37
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