SQL for nested WITH CLAUSE - RESULTS OFFSET in Oracle 19c - oracle19c

Question

Please suggest a way to implement nesting of (temp - results - select) as shown below?
I see that oracle 19c does not allow nesting of WITH clause.
with temp2 as
(
with temp1 as
(
__
__
),
results(..fields..) as
(
select ..<calc part>.. from temp1, results where __
)
select ..<calc part>.. from temp1 join results where __
),
results(..fields..) as
(
select ..<calc part>.. from temp2, results where __
)
select ..<calc part>.. from temp2 join results where __
For instance:
DB Fiddle
I need to calculate CALC3 in similar recursive way as of CALC
CREATE TABLE TEST ( DT DATE, NAME VARCHAR2(10), VALUE NUMBER(10,3));
insert into TEST values ( to_date( '01-jan-2021'), 'apple', 198.95 );
insert into TEST values ( to_date( '02-jan-2021'), 'apple', 6.15 );
insert into TEST values ( to_date( '03-jan-2021'), 'apple', 4.65 );
insert into TEST values ( to_date( '06-jan-2021'), 'apple', 20.85 );
insert into TEST values ( to_date( '01-jan-2021'), 'banana', 80.5 );
insert into TEST values ( to_date( '02-jan-2021'), 'banana', 9.5 );
insert into TEST values ( to_date( '03-jan-2021'), 'banana', 31.65 );
--Existing working code -
with t as
( select
test.*,
row_number() over ( partition by name order by dt ) as seq
from test
),
results(name, dt, value, calc ,seq) as
(
select name, dt, value, value/5 calc, seq
from t
where seq = 1
union all
select t.name, t.dt, t.value, ( 4 * results.calc + t.value ) / 5, t.seq
from t, results
where t.seq - 1 = results.seq
and t.name = results.name
)
select results.*, calc*3 as calc2 -- Some xyz complex logic as calc2
from results
order by name, seq;
Desired output:
CALC3 - grouped by name and dt -
((CALC3 of prev day record * 4) + CALC2 of current record )/ 5
i.e for APPLE
for 1-jan-21, CALC = ((0*4)+119.37)/5 = 23.87 -------> since it is 1st record, have taken 0 as CALC3 of prev day record
for 2-jan-21, CALC = ((23.87*4)+99.19)/5= 115.33 -----> prev CALC3 is considered from 1-jan-21 - 23.87 and 99.19 from current row
for 3-jan-21, CALC = ((115.33*4)+82.14)/5= 477.76 and so on
For BANANA
1-jan-21, CALC = ((0*4)+48.30)/5=9.66
1-jan-21, CALC = ((9.66*4)+44.34)/5=47.51
etc

Answer 1

You do not need to, you can just do it all in one level:
with temp1(...fields...) as
(
__
__
__
),
results1(...fields...) as
(
select ...<calc part>... from temp1 where __
),
temp2( ...fields...) as
(
select ...<calc part>... from temp1 join results1 where __
),
results2(...fields...) as
(
select ...<calc part>... from temp2 where __
)
select ...<calc part>... from temp2 join results2 where __
For your actual problem, you can use a MODEL clause:
SELECT dt,
name,
amount,
calc,
seq,
calc2,
calc3
FROM (
SELECT t.*,
ROW_NUMBER() OVER (PARTITION BY name ORDER BY dt) AS seq
FROM test t
)
MODEL
PARTITION BY (name)
DIMENSION BY (seq)
MEASURES ( dt, amount, 0 AS calc, 0 AS calc2, 0 as calc3)
RULES (
calc[1] = amount[1]/5,
calc[seq>1] = (amount[cv(seq)] + 4*calc[cv(seq)-1])/5,
calc2[seq] = 3*calc[cv(seq)],
calc3[1] = calc2[1]/5,
calc3[seq>1] = (calc2[cv(seq)] + 4*calc3[cv(seq)-1])/5
)
Which outputs:
DT
NAME
AMOUNT
CALC
SEQ
CALC2
CALC3
01-JAN-21
banana
80.5
16.1
1
48.3
9.66
02-JAN-21
banana
9.5
14.78
2
44.34
16.596
03-JAN-21
banana
31.65
18.154
3
54.462
24.1692
01-JAN-21
apple
198.95
39.79
1
119.37
23.874
02-JAN-21
apple
6.15
33.062
2
99.186
38.9364
03-JAN-21
apple
4.65
27.3796
3
82.1388
47.57688
06-JAN-21
apple
20.85
26.07368
4
78.22104
53.705712
db<>fiddle here