SQL Query calculating two additional columns - sql

Question

I have a table which gets populated daily with database size. I need to modify the query where I can calculate daily growth and weekly growth.
select * from sys.dbsize
where SNAP_TIME > sysdate -3
order by SNAP_TIME
Current Output
I would like to add two additional columns which would be
Daily Growth (DB_SIZE sysdate - DB_SIZE (sysdate -1))
Weekly Growth (DB_SIZE sysdate - DB_SIZE (sysdate -7))
Need some help constructing the SQL for those two additional columns. Any help will be greatly appreciated.
Thanks,

Answer 1

One option is to use LAG analytic function to calculate daily growth and correlated subquery (within the SELECT statement) for weekly growth.
For example:
SQL> with dbsize (snap_time, db_size) as
2 (select sysdate - 8, 100 from dual union all
3 select sysdate - 7, 110 from dual union all
4 select sysdate - 6, 105 from dual union all
5 select sysdate - 5, 120 from dual union all
6 select sysdate - 4, 130 from dual union all
7 select sysdate - 3, 130 from dual union all
8 select sysdate - 2, 142 from dual union all
9 select sysdate - 1, 144 from dual union all
10 select sysdate - 0, 150 from dual
11 )
12 select
13 a.snap_time,
14 a.db_size,
15 a.db_size - lag(a.db_size) over (order by a.snap_time) daily_growth,
16 --
17 db_size - (select db_size from dbsize b
18 where trunc(b.snap_time) = trunc(a.snap_time) - 7
19 ) weekly_growth
20 from dbsize a
21 order by a.snap_time;
SNAP_TIME DB_SIZE DAILY_GROWTH WEEKLY_GROWTH
------------------- ---------- ------------ -------------
24.08.2020 21:52:20 100
25.08.2020 21:52:20 110 10
26.08.2020 21:52:20 105 -5
27.08.2020 21:52:20 120 15
28.08.2020 21:52:20 130 10
29.08.2020 21:52:20 130 0
30.08.2020 21:52:20 142 12
31.08.2020 21:52:20 144 2 44
01.09.2020 21:52:20 150 6 40
9 rows selected.
SQL>

Answer 2

I would recommend lag() for both columns:
select s.*,
(dbsize - dbsize_1) as daily_growth,
(dbsize - dbsize_7) as weekly_growth
from (select s.*,
lag(dbsize) over (order by snap_time) as dbsize_1,
lag(dbsize, 7) over (order by snap_time) as dbsize_7
from sys.dbsize
) s
where SNAP_TIME > sysdate -3
order by SNAP_TIME;
If you don't have a snapshot each day, you can handle this with a window frame:
select s.*,
(dbsize - dbsize_1) as daily_growth,
(dbsize - dbsize_7) as weekly_growth
from (select s.*,
max(dbsize) over (order by trunc(snap_time) range between interval '1' day preceding and interval '1' second preceding) as dbsize_1,
lag(dbsize, 7) over (order by trunc(snap_time) range between '7' day preceding and interval '6 1' day to hour) as dbsize_7
from sys.dbsize
) s
where SNAP_TIME > sysdate - 3
order by SNAP_TIME;

Answer 3

If there is always is one record per day, you can use lag():
select
snap_time
db_size,
db_size - lag(db_size, 1) over(order by snap_time) daily_growth,
db_size - lag(db_size, 7) over(order by snap_time) weekly_growth
from sys.db.size
order by snap_time
This actually looks 1 row back and 7 rows back. If there are missing dates, or multiple records per day, then you could average the snap size by day, and use a window range in the window function:
select
trunc(snap_time) snap_day,
avg(db_size) avg_db_size,
avg(db_size) - avg(db_size) over(
order by trunc(snap_time)
range between interval '1' day preceding and interval '1' day preceding
) daily_growth,
avg(db_size) - avg(db_size) over(
order by trunc(snap_time)
range between interval '7' day preceding and interval '7' day preceding
) weekly_growth
from sys.db.size
group by trunc(snap_time)
order by trunc(snap_time)
If you want the results for the last 3 days only, you can turn any of the two above queries to subqueries, and filter in the outer query:
select *
from ( ... ) t
where snap_time > sysdate - 3 -- or: snap_day > trunc(sysdate) - 3