How to group-by in Oracle - sql

Question

I have a table like [Original] in below.
I want to sum by group-by field like [result].
Does anyone have an idea to make this query?
Thank you in advance for your help.
WITH t1 as (
SELECT 1 AS ID, 'A' AS FIELD, 1 AS VAL FROM dual
UNION SELECT 2 AS ID, 'A' AS FIELD, 2 AS VAL FROM dual
UNION SELECT 3 AS ID, 'A' AS FIELD, 1 AS VAL FROM dual
UNION SELECT 4 AS ID, 'B' AS FIELD, 2 AS VAL FROM dual
UNION SELECT 5 AS ID, 'B' AS FIELD, 2 AS VAL FROM dual
UNION SELECT 6 AS ID, 'B' AS FIELD, 1 AS VAL FROM dual
UNION SELECT 7 AS ID, 'A' AS FIELD, 3 AS VAL FROM dual
UNION SELECT 8 AS ID, 'A' AS FIELD, 2 AS VAL FROM dual
UNION SELECT 9 AS ID, 'A' AS FIELD, 1 AS VAL FROM dual
)
SELECT *
FROM t1
[Original Data]
ID FIELD VAL
1 A 1
2 A 2
3 A 1
4 B 2
5 B 2
6 B 1
7 A 3
8 A 2
9 A 1
[Result]
ID FIELD VAL
1 A 4
4 B 5
7 A 6

Answer 1

This is island and gap issue and you can use analytical function as follows:
SQL> WITH t1 as (
2 SELECT 1 AS ID, 'A' AS FIELD, 1 AS VAL FROM dual
3 UNION SELECT 2 AS ID, 'A' AS FIELD, 2 AS VAL FROM dual
4 UNION SELECT 3 AS ID, 'A' AS FIELD, 1 AS VAL FROM dual
5 UNION SELECT 4 AS ID, 'B' AS FIELD, 2 AS VAL FROM dual
6 UNION SELECT 5 AS ID, 'B' AS FIELD, 2 AS VAL FROM dual
7 UNION SELECT 6 AS ID, 'B' AS FIELD, 1 AS VAL FROM dual
8 UNION SELECT 7 AS ID, 'A' AS FIELD, 3 AS VAL FROM dual
9 UNION SELECT 8 AS ID, 'A' AS FIELD, 2 AS VAL FROM dual
10 UNION SELECT 9 AS ID, 'A' AS FIELD, 1 AS VAL FROM dual
11 )
12 SELECT MIN(ID) AS ID, FIELD, SUM(VAL)
13 FROM (SELECT T1.*,
14 SUM(CASE WHEN LAG_FIELD = FIELD THEN 0 ELSE 1 END)
15 OVER (ORDER BY ID) AS SM
16 FROM (SELECT T1.*,
17 LAG(FIELD) OVER (ORDER BY ID) AS LAG_FIELD
18 FROM t1
19 ) T1
20 )
21 GROUP BY FIELD, SM
22 ORDER BY 1;
ID F SUM(VAL)
---------- - ----------
1 A 4
4 B 5
7 A 6
SQL>

Answer 2

This is indeed a gaps-and-islands problem. I think the simplest approach here is to use the difference between row numbers to identify groups of adjacent rows:
select min(id) as id, field, sum(val) as val
from (
select t1.*,
row_number() over(order by id) rn1,
row_number() over(partition by field order by id) rn2
from t1
) t
group by field, rn1 - rn2
order by min(id)
If id is always incrementing without gaps, this is even simpler:
select min(id) as id, field, sum(val) as val
from (
select t1.*,
row_number() over(partition by field order by id) rn
from t1
) t
group by field, id - rn
order by min(id)

Answer 3

From Oracle 12, you can do it quite simply using MATCH_RECOGNIZE:
WITH t1 as (
SELECT 1 AS ID, 'A' AS FIELD, 1 AS VAL FROM dual
UNION SELECT 2 AS ID, 'A' AS FIELD, 2 AS VAL FROM dual
UNION SELECT 3 AS ID, 'A' AS FIELD, 1 AS VAL FROM dual
UNION SELECT 4 AS ID, 'B' AS FIELD, 2 AS VAL FROM dual
UNION SELECT 5 AS ID, 'B' AS FIELD, 2 AS VAL FROM dual
UNION SELECT 6 AS ID, 'B' AS FIELD, 1 AS VAL FROM dual
UNION SELECT 7 AS ID, 'A' AS FIELD, 3 AS VAL FROM dual
UNION SELECT 8 AS ID, 'A' AS FIELD, 2 AS VAL FROM dual
UNION SELECT 9 AS ID, 'A' AS FIELD, 1 AS VAL FROM dual
)
SELECT *
FROM t1
MATCH_RECOGNIZE (
ORDER BY id
MEASURES
FIRST( id ) AS id,
FIRST( field ) AS field,
SUM( val ) AS total
ONE ROW PER MATCH
PATTERN( same_field+ )
DEFINE same_field AS FIRST(field) = field
)
Which outputs:
ID | FIELD | TOTAL
-: | :---- | ----:
1 | A | 4
4 | B | 5
7 | A | 6
db<>fiddle here