Get last date of modification in database by value - sql

Question

How it is possible to get - when was the last change (by date) - in
this table:
id
date
value
1
01.01.2021
0.0
1
02.01.2021
10.0
1
03.01.2021
15.0
1
04.01.2021
25.0
1
05.01.2021
25.0
1
06.01.2021
25.0
Of course I could use clause where and it will works, but i have a lot of rows and for some i don't now exactly day when this happend.
The resault should be:
id
date
value
1
04.01.2021
25.0

Answer 1

Try this one:
with mytable as (
select 1 as id, date '2021-01-01' as date, 0 as value union all
select 1, date '2021-01-02', 10 union all
select 1, date '2021-01-03', 15 union all
select 1, date '2021-01-04', 25 union all
select 1, date '2021-01-05', 25 union all
select 1, date '2021-01-06', 25
)
select id, array_agg(struct(date, value) order by last_change_date desc limit 1)[offset(0)].*
from (
select *, if(value != lag(value) over (partition by id order by date), date, null) as last_change_date
from mytable
)
group by id

Answer 2

in this scenario I would be using two field in my database "created_at and updated_at" with the type as "timestamp". You may simply fetch your records using OrderBY "updated_at" field.

Answer 3

see what this gives you:
SELECT MAX(date) OVER (PARTITION BY(value)) AS lastChange
FROM Table
WHERE id = 1

Answer 4

The following query and reproducible example on db-fiddle works. I've also included some additional test records.
CREATE TABLE my_data (
`id` INTEGER,
`date` date,
`value` INTEGER
);
INSERT INTO my_data
(`id`, `date`, `value`)
VALUES
('1', '01.01.2021', '0.0'),
('1', '02.01.2021', '10.0'),
('1', '03.01.2021', '15.0'),
('1', '04.01.2021', '25.0'),
('1', '05.01.2021', '25.0'),
('1', '06.01.2021', '25.0'),
('2', '05.01.2021', '25.0'),
('2', '06.01.2021', '23.0'),
('3', '03.01.2021', '15.0'),
('3', '04.01.2021', '25.0'),
('3', '05.01.2021', '17.0'),
('3', '06.01.2021', '17.0');
Query #1
SELECT
id,
date,
value
FROM (
SELECT
*,
row_number() over (partition by id order by date desc) as id_rank
FROM (
SELECT
id,
m1.date,
m1.value,
rank() over (partition by id,m1.value order by date asc) as id_value_rank,
CASE
WHEN (m1.date = (max(m1.date) over (partition by id,m1.value ))) THEN 1
ELSE 0
END AS is_max_date_for_group,
CASE
WHEN (m1.date = (max(m1.date) over (partition by id ))) THEN 1
ELSE 0
END AS is_max_date_for_id
from
my_data m1
) m2
WHERE (m2.is_max_date_for_group = m2.is_max_date_for_id and is_max_date_for_group <> 0 and id_value_rank=1) or (id_value_rank=1 and is_max_date_for_id=0)
) t
where t.id_rank=1
order by id, date, value;
id
date
value
1
04.01.2021
25
2
06.01.2021
23
3
05.01.2021
17
View on DB Fiddle

Answer 5

I actually find that the simplest method is to enumerate the rows by id/date and by id/date/value in descending order. These are the same for the last group . . . and the rest is aggregation:
select id, min(date), value
from (select t.*,
row_number() over (partition by id order by date desc) as seqnum,
row_number() over (partition by id, value order by date desc) as seqnum_2
from t
) t
where seqnum = seqnum_2
group by id;
If you use lag(), I would recommend using qualify for performance:
select t.*
from (select t.*
from t
qualify lag(value) over (partition by id order by date) <> value or
lag(value) over (partition by id order by date) is null
) t
qualify row_number() over (partition by id order by date desc) = 1;
Note: Both of these work if the value is the same for all rows. Other methods may not work in that situation.