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Calculate difference between values with MySQL (PV)

I'm a bit stuck at my Bachelor's thesis and hope you can help me. In order to evaluate a photovoltaic system I need to calculate the difference between total energy amounts. These are automatically updated in a MySQL-table with a timestamp, but without an id-number. I need to get the delta/difference between those energy amounts automatically as a extra column to visualize it in Grafana.

******************************************
Timestamp            | SB1_AC_Total       | Needed information (delta)
******************************************
2020-06-24 09:32:45  | 11.326.302         |   23

2020-06-24 09:32:02  | 11.326.279         |   22

2020-06-24 09:31:20  | 11.326.257         |   ...

This list goes on for weeks. I really hope you can help me, because I have no idea and it is the first time I work with MySQL.

over 4 years ago · Santiago Trujillo
3 Respuestas
Responde la pregunta

0

Consider following code and use it as example:

create table test (time timestamp, total int);

insert into test values (current_timestamp() - interval 2 day, 100);
insert into test values (current_timestamp() - interval 1 day, 120);
insert into test values (current_timestamp(),                  125);

select i.*, i.total - t_outer.total as diff 
from (select t.*, (select max(time) 
                   from test t1 where t1.time < t.time) as last_timestamp
      from test t
) as i 
left join test t_outer
on i.last_timestamp = t_outer.time;

As result you get something like:

time    total   last_timestamp  diff
"2020-06-23 10:58:21"   120 "2020-06-22 10:58:19"   20
"2020-06-24 10:58:22"   125 "2020-06-23 10:58:21"   5
"2020-06-22 10:58:19"   100 NULL    NULL

EDIT:

If you want to put the diff value into a new table column, you can do it like this:

alter table test add column diff int default null;

create table select_bkp as -- you know the code below already
select i.time, i.total - t_outer.total as diff 
from (select t.*, (select max(time) 
                   from test t1 where t1.time < t.time) as last_timestamp
      from test t
) as i 
left join test t_outer
on i.last_timestamp = t_outer.time;

SET SQL_SAFE_UPDATES = 0; -- so all rows can be updated at once

update test t
join select_bkp b
  on t.time = b.time
set t.diff = b.diff;

SET SQL_SAFE_UPDATES = 1; -- enable secure updates
over 4 years ago · Santiago Trujillo Denunciar

0

I was able to answer my question with the following code:


ALTER TABLE TABLENAME  ADD SB1_AC_GES_DIFF INT;
DELIMITER $$
CREATE TRIGGER TABLENAME_Trigger
    BEFORE INSERT
    ON TABLENAME FOR EACH ROW
BEGIN
   DECLARE SB1_AC_GES_old INT;  
  SELECT max(SB1_AC_GES) INTO SB1_AC_GES_old FROM TABLENAME;
   SET NEW.SB1_AC_GES_Diff = New.SB1_AC_GES - SB1_AC_GES_old;
END$$    
DELIMITER ;

over 4 years ago · Santiago Trujillo Denunciar

0

How you approach this depends on your skill set with various tools. I, for example, am old in 'C' but green in SQL.

Now: Assuming that you have a database already, and want to create a new one with the extra field in it, this is broadly how I would approach it.

Create a new empty database with the fields you want in it.

open both databases 

Read all the old database into a program coded in whatever language you are happy with. 
Typically you will access this by executing an SQL statement like

SELECT * from table1 order by timestamp

In general you can now step through the data base line by line. 
The algorithm you want is this

set a variable 'delta' to zero
set a variable 'flag to zero
while(more lines)
    read 'timestamp'
    read 'SB1_AC_Total'
    write 'timestamp', 'SB1_AC_Total', and 'delta' to the new database using sql INSERT command.
    If 'flag' is not zero, set 'delta' to ('SB1_AC_Total' - 'old_data');  endif
    set 'flag' to 1
    set a variable 'old_data' to 'SB1_AC_Total'
endwhile 

This will duplicate the database with a new column whose value is the current value minus the previous value.

I suspect there is an SQL way using two selects to generate two temporary tables and then reading from one and the other and combining the result into a third, but its beyond me.

over 4 years ago · Santiago Trujillo Denunciar
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