Reputation: 190
I have an SQL table with "lines" representing elements of chained lists. I could for example have the following records:
(id, previous_id)
------------------
(1, NULL)
(2, NULL)
(3, 2)
(4, 3)
(5, NULL)
(6, 4)
(7, 5)
We have 3 lists in this table:
(1,)
(2,3,4,6)
(5,7)
I would like to find the last element of each list and the number of elements in the list. The query I am looking for would output:
last, len
1, 1
6, 4
7, 2
Is this possible in SQL?
Upvotes: 1
Views: 886
Reputation: 1
Here is an implementation in Microsoft SQL Server 2016 db<>fiddle
WITH chain
AS (SELECT l.id AS [first],
l.id AS [last],
1 AS [len]
FROM lines AS l
WHERE l.previous_id IS NULL
UNION ALL
SELECT c.[first],
l.id,
c.[len] + 1 AS [len]
FROM chain AS c
JOIN lines AS l ON l.previous_id = c.[last]),
result
AS (SELECT DISTINCT
c.[first],
c.[last],
c.[len],
ROW_NUMBER() OVER(PARTITION BY c.[first] ORDER BY c.[len] DESC) AS rn
FROM chain as c)
SELECT r.[first],
r.[last],
r.[len]
FROM result AS r
WHERE r.rn = 1
ORDER BY r.[first];
Upvotes: 0
Reputation: 1270573
You can use a recursive CTE:
with recursive cte as (
select l.previous_id as id, id as last
from lines l
where not exists (select 1 from lines l2 where l2.previous_id = l.id)
union all
select l.previous_id, cte.last
from cte join
lines l
on cte.id = l.id
)
select cte.last, count(*)
from cte
group by cte.last;
Here is a db<>fiddle.
Upvotes: 1
Reputation: 658092
WITH RECURSIVE cte AS (
SELECT id AS first, id AS last, 1 as len
FROM lines
WHERE previous_id IS NULL
UNION ALL
SELECT c.first, l.id, len + 1
FROM cte c
JOIN lines l ON l.previous_id = c.last
)
SELECT DISTINCT ON (first)
last, len -- , first -- also?
FROM cte
ORDER BY first, len DESC;
db<>fiddle here
Produces your result exactly.
If yo also want the first element like your title states, that's readily available.
Upvotes: 0