按月获取最小值和最大值组之间的差异

Question

I've a database with two columns.

Column 1: named as datum (datetime format)
Column 2: named as Energie (decimal format).

2019-01-01 14:17:07;3419.699951
2019-01-01 14:48:38;3419.800049
2019-01-01 15:12:35;3419.900146
2019-01-01 15:34:26;3420.100098
2019-01-01 15:58:10;3420.300049
2019-01-01 16:21:36;3420.5
2019-01-01 17:20:03;3420.699951
2019-01-01 17:32:11;3420.800049
2019-01-01 17:56:17;3421
2019-01-02 04:40:50;3421.100098
2019-01-02 05:46:28;3421.299805
2019-01-02 05:55:49;3421.400146
2019-01-02 06:41:54;3421.599854
2019-01-02 07:03:16;3421.700195
2019-01-02 07:26:59;3421.899902
2019-01-02 07:39:02;3422
2019-01-02 07:50:57;3422.100098
2019-01-02 08:50:53;3422.300049
2019-01-02 09:12:43;3422.5
2019-01-02 09:24:47;3422.600098

I would like to calculate the difference between the min and max value in column 2 per month and group the results per month.

Till this far I have the following SQL statement what resulted in difference per day.

select  cast(datum as date) AS datum, (max(energie) - min(energie)) as Energie_MWh
from    `01` as p
group by 
cast(datum as date)

Answer 1

If you are getting data for each day and you want it for month then you can use DATE_FORMAT(cast(datum as date), "%Y%m") instead of cast(datum as date)

Answer 2

These lines work for MySQL:

select DATE_FORMAT(datum, "%Y%m")  as date_month,
   max(energy)                 as max_engery,
   min(energy)                 as min_energy,
   (max(energy) - min(energy)) as diff
from energy_table
group by date_month;

But I wonder if this is a real problem for engergy usage?

And it will cause a performance problem if the table is large.

Answer 3

MySQL has the very convenient YEAR_MONTH option for extract() , so you could use:

select extract(year_month from datum) as yyyymm,
       (max(energie) - min(energie)) as Energie_MWh
from `01` as p
group by yyyymm;