Extract difference in consecutive values for MySQL 5.7

Question

Name	Date	Hours	Count
Mills	2022-07-17	23	12
Mills	2022-07-18	00	15
Mills	2022-07-18	01	20
Mills	2022-07-18	02	22
Mills	2022-07-18	03	25
Mills	2022-07-18	04	20
Mills	2022-07-18	05	22
Mills	2022-07-18	06	25
MIKE	2022-07-18	00	15
MIKE	2022-07-18	01	20
MIKE	2022-07-18	02	22
MIKE	2022-07-18	03	25
MIKE	2022-07-18	04	20

My current input table stores information for counts recorded in each hour of the day consecutively. I need to extract the difference in values for consecutive counts but I'm having trouble doing it since I'm forced to use MySQL 5.7.

I have written the query as follows:

SET @cnt := 0;
SELECT Name, Date, Hours, Count, (@cnt := @cnt - Count) AS DiffCount
FROM Hourly
ORDER BY Date;

which is not giving exact results.

I expect to have the following output:

Name	Date	Hours	Count	Diff
Mills	2022-07-17	23	12	0
Mills	2022-07-18	00	15	3
Mills	2022-07-18	01	20	5
Mills	2022-07-18	02	22	2
Mills	2022-07-18	03	25	3
Mills	2022-07-18	04	20	5
Mills	2022-07-18	05	22	2
Mills	2022-07-18	06	25	3
MIKE	2022-07-18	00	15	0
MIKE	2022-07-18	01	20	5
MIKE	2022-07-18	02	22	2
MIKE	2022-07-18	03	25	3
MIKE	2022-07-18	04	20	5
MIKE	2022-07-18	05	22	2
MIKE	2022-07-18	06	25	3

Please suggest what I'm missing.

Answer 1

Try the following:

SET @count=(select Count_ from Hourly order by Name,Date_,Hours LIMIT 1);
Set @Name=(select Name from Hourly order by Name,Date_,Hours LIMIT 1);

select  Name,Date_,Hours,Count_,
ABS(curr_count-lag_count) as DiffCount
From
(
  select Name,Date_,Hours,Count_, 
  Case When @Name = Name Then @count Else Count_ End as lag_count
  , @count:=Count_ curr_count, @Name:=Name
  From Hourly order by Name,Date_,Hours
) D
Order By Name,Date_,Hours;

See a demo from db-fiddle.

Answer 2

In MySQL 5.7 you can update your variable inline in order to contain your updated value of " Count ". Since when the value of " Name " changes you need to reset your variable, you can use another variable that contains the previous " Name " value. Then you use an IF function to check:

• if your previous name is equal to your current name
• then take the difference in counts
• else assign 0

It would be combined with the ABS function, which applies the absolute value on the difference.

SET @cnt  := NULL;
SET @name := NULL;

SELECT Date,
       Hours,
       ABS(IF(@name = Name, 
              @cnt := @cnt - Count,
              0)                    ) AS DiffCount,
       (@name := Name)                AS Name,
       (@cnt := Count)                AS Count
FROM tab
ORDER BY Name DESC,
         Date, 
         Hours;

Check the demo here .

---

In MySQL 8.0 you can use a window fuction like LAG to get your output smoothly. It would be combined with:

• ABS that applies the absolute value on the difference,
• COALESCE employed for removing the first null value.

SELECT *, 
       COALESCE(ABS(
           Count - LAG(Count) OVER(
                       PARTITION BY Name
                       ORDER     BY Date, Hours
                   )
       ), 0) AS Diff 
FROM tab
ORDER BY Name DESC, 
         Date, 
         Hours

Check the demo here .