使用 SQL 数据库查询以不同的时间间隔减去相同的列值

Question

在 MySQL 中，我想根据另一列“时间戳”减去不同时间间隔的列值。

table structure is : 
id | generator_id | timestamp             | generated_value
1  |  1           | 2019-05-27 06:55:20   | 123456     
2  |  1           | 2019-05-27 07:55:20   | 234566       
3  |  1           | 2019-05-27 08:55:20   | 333456       
..
..    
20  |  1           | 2019-05-27 19:55:20   | 9876908       

从上表中我想获取 generate_value 列值，它应该是第一个时间戳的日期和日期的最后一个值的时间戳的差异。

在上面的示例中，我正在查找应该输出为 9,753,452 (9876908 - 123456) 的查询。

一般来说，为了获取我在查询下面使用的第一个值和最后一个值的单个记录

// Below will give me end day value
SELECT * FROM generator_meters where generator_id=1 and timestamp like '2019-05-27%' order by timestamp desc limit 1 ;

//this will give me last day value 
SELECT * FROM generator_meters where generator_id=1 and timestamp like '2019-05-27%' order by timestamp  limit 1 ;

问题是我应该如何通过从一天的最后一个值减去一天的第一个值来获得最终的生成值。

Expected Output 
 generator_id | generated_value
  1           | 9753452     

提前致谢！！

Answer 1

在您的示例中，值变得越来越大。如果保证是这样，您可以使用

select max(generated_value) - min(generated_value) as result
from sun_electric.generator_meters
where generator_id = 1
and date(timestamp) = date '2019-05-27';

或者对于多个 ID：

select generator_id, max(generated_value) - min(generated_value) as result
from sun_electric.generator_meters
and date(timestamp) = date '2019-05-27'
group by generator_id
order by generator_id;

如果值不是升序的，那么您可以对 ID 1 使用以下查询：

select last_row.generated_value - first_row.generated_value as result
from 
(
  select * 
  from sun_electric.generator_meters
  where generator_id = 1
  and date(timestamp) = date '2019-05-27'
  order by timestamp
  limit 1
) first_row
cross join
(
  select * 
  from sun_electric.generator_meters
  where generator_id = 1
  and date(timestamp) = date '2019-05-27'
  order by timestamp desc
  limit 1
) last_row;

这是获取多个 ID 的结果的一种方法：

select
  minmax.generator_id,
  (
    select generated_value 
    from sun_electric.generator_meters gm
    where gm.generator_id = minmax.generator_id
    and gm.timestamp = minmax.max_ts
  ) -
  (
    select generated_value 
    from sun_electric.generator_meters gm
    where gm.generator_id = minmax.generator_id
    and gm.timestamp = minmax.min_ts
  ) as result
from
(
  select generator_id, min(timestamp) as min_ts, max(timestamp) as max_ts
  from sun_electric.generator_meters
  where date(timestamp) = date '2019-05-27'
  group by generator_id
) minmax
order by minmax.generator_id;

如果您更喜欢这个，您也可以将子查询移动到 from 子句并加入它们。另一种方法是使用从 MySQL 8 开始可用的窗口函数。

Answer 2

以下脚本将返回过滤后的 ID 和日期的预期结果-

SELECT generator_id,CAST(timestamp AS DATE) ,
(
    SELECT generated_value 
    FROM sun_electric.generator_meters B 
    WHERE timestamp = max(timestamp)
)
-
(
    SELECT generated_value 
    FROM sun_electric.generator_meters B 
    WHERE timestamp = min(timestamp)
) AS Diff
FROM sun_electric.generator_meters 
WHERE generator_id = 1
AND CAST(timestamp AS DATE) = '2019-05-27'
GROUP BY generator_id,CAST(timestamp AS DATE) ; 

如果您希望使用 GROUP BY ID 和日期获得相同的结果，只需删除过滤器，如下所示-

SELECT generator_id,CAST(timestamp AS DATE) ,
(
    SELECT generated_value 
    FROM sun_electric.generator_meters B 
    WHERE timestamp = max(timestamp)
)
-
(
    SELECT generated_value 
    FROM sun_electric.generator_meters B 
    WHERE timestamp = min(timestamp)
) AS Diff
FROM sun_electric.generator_meters 
GROUP BY generator_id,CAST(timestamp AS DATE) ;