Putting two rows together on the same row in Oracle - sql

Question

Let's take as example this table, which associate a product to the number
of stars he received from customer's feedbacks (it happens to be ordered by number of stars):
| ProductID | Stars |
|-----------|---------|
| 23 | 10 |
| 12 | 10 |
| 17 | 9 |
| 5 | 8 |
| 20 | 8 |
| 18 | 7 |
How would I select (showing them on the same row) the IDs of the products
in pairs?
Like this:
| Product1 | Product2 |
|-----------|-------------|
| 23 | 12 |
| 17 | 5 |
| 20 | 18 |
Or like this:
| Products |
|------------------|
| 23 12 |
| 17 5 |
| 20 18 |

Answer 1

Use listagg to produce option2
select stars, listagg(ProductID, ' ') within group (order by ProductID) as Products
from Table1
group by Stars

Answer 2

It is not clear why in your output you picked 5 before 20; remember that rows in a table are NOT ORDERED. In my solution I order by stars and then by productid; if your rows are ordered by something else as well, you may use that instead.
You may change 2 to 7 in the divisions by 2 if you want to group 7 values at a time.
This is offered only to show that this CAN be done in Oracle SQL (in the database). It SHOULDN'T, but that's for you to decide.
with
inputs ( productid, stars) as (
select 23, 10 from dual union all
select 12, 10 from dual union all
select 17, 9 from dual union all
select 5, 8 from dual union all
select 20, 8 from dual union all
select 18, 7 from dual
)
-- end of test data, solution begins below
select listagg(productid, ' ') within group (order by rn) as result
from ( select productid, stars,
row_number() over (order by stars desc, productid desc) as rn
from inputs
)
group by ceil(rn/2)
order by ceil(rn/2)
;
RESULT
------
23 12
17 20
5 18

Answer 3

WITH t
AS (SELECT 23 product_id, 10 stars FROM DUAL
UNION ALL
SELECT 12, 10 FROM DUAL
UNION ALL
SELECT 17, 9 FROM DUAL
UNION ALL
SELECT 5, 8 FROM DUAL
UNION ALL
SELECT 20, 8 FROM DUAL
UNION ALL
SELECT 18, 7 FROM DUAL),
t2
AS ( SELECT product_id,
stars,
ROW_NUMBER () OVER (ORDER BY stars DESC)
+ MOD (ROW_NUMBER () OVER (ORDER BY stars DESC), 2)
grp
FROM t
ORDER BY stars DESC)
SELECT LISTAGG (product_id, ' ') WITHIN GROUP (ORDER BY stars DESC, ROWNUM)
AS product_id
FROM t2
GROUP BY grp
result
23 12
17 5
20 18

Answer 4

On oracle you can use a function call listagg which works sort of like join in C# (and other languages.) For it to work you need to have something to group buy -- it seems you want to group by every other when ordered by a column. First make a row number for the column and then group by that div 2
so
SELECT LISTAGG(ProductID, ' ') WITHIN GROUP (ORDER BY ProductID) AS ProductList
FROM (
SELECT ProductID, FLOOR((ROW_NUMBER() OVER (ORDER BY Stars)+1)/2) as GroupByMe FROM Table
) X
GROUP BY GroupByMe
If you want to do every 3, just /3 instead.
More interesting (because more people want to see it) is getting a list for every amount of stars, that looks like this:
SELECT Stars as StarCount, LISTAGG(ProductID, ' ') WITHIN GROUP (ORDER BY ProductID) As ProductList
FROM Table
GROUP BY Stars
of course this looks like a prior answer because this is what people would expect you would want.