Cube.dev - how to have the parameter "today date" in rolling window? - postgresql

Question

I have a PostgreSQL database in which users can order from a start date to an end date.
I want to know, for each day, how many users would be able to order.
Let's make an example, given:
id
from
to
A
2023-01-01
2023-01-03
B
2023-01-02
2023-01-07
C
2023-01-02
2023-01-09
D
2023-01-10
2023-01-12
For the first two weeks (let's suppose that the 01/01/2023 were on Monday), I would have (the third column is not needed):
day
n_of_users_that_can_order
who
2023-01-01
1
A
2023-01-02
3
A, B, C
2023-01-03
3
A, B, C
2023-01-04
2
B, C
2023-01-05
2
B, C
2023-01-06
2
B, C
2023-01-07
2
B, C
2023-01-08
1
C
2023-01-09
1
C
2023-01-10
1
D
2023-01-11
1
D
2023-01-12
0
2023-01-13
0
2023-01-14
0
My end result should be the above table aggregated per week:
week
total
2023-01-01 to 2023-01-07
15
2023-01-08 to 2023-01-14
4
I don't know how to do it with cube.dev, this is the idea I had for now:
ube(`Users`, {
sql: `SELECT * FROM public.users`,
measures: {
usersPerDayThatCanOrder: {
type: `count`,
sql: `id`,
rollingWindow: {
trailing: `1 day`,
offset: `start`
},
filters: [
{ sql: `${CUBE}.can_order_from >= ${TODAY} AND ${CUBE}.can_order_to <= ${TODAY}` }, // NOTE: today doesn't exist
]
}
},
dimensions: {
id: {
sql: `id`,
type: `number`,
primaryKey: true
},
canOrderFrom: {
sql: `can_order_from`,
type: `time`
},
canOrderTo: {
sql: `can_order_to`,
type: `time`
},
},
dataSource: `default`
});
But it does not work because I do not know how to have the real ${TODAY} value and also how to aggregate per week.

Answer 1

Here is a sql query which provides your expected result assuming that your table is named test :
SELECT d.week AS "week start"
, (d.week + interval '6 days') :: date AS "week end"
, sum( upper(daterange(d.week, (d.week + interval '1 week') :: date, '[)') * daterange(t."from", t."to", '[]'))
- lower(daterange(d.week, (d.week + interval '1 week') :: date, '[)') * daterange(t."from", t."to", '[]'))
)
FROM
( SELECT generate_series(min(date_trunc('week', "from")), max("to"), interval '1 week') :: date AS week
FROM test
) AS d
INNER JOIN test AS t
ON daterange(d.week, (d.week + interval '1 week') :: date, '[)') && daterange(t."from", t."to", '[]')
GROUP BY d.week
ORDER BY d.week
The subquery calculates the weeks start date covered by table test
The INNER JOIN clause intersects the weeks with the user date ranges
Then rows are concatenated by weeks and the number of days are summed for all the users
By the way, the 2023/01/01 seems not to be Monday but Sunday.
Result :
week start
week end
sum
2022-12-26
2023-01-01
1
2023-01-02
2023-01-08
15
2023-01-09
2023-01-15
4
see test result in dbfiddle