sql: first row after the last row with a property - sql

Question

I would like to write a query that returns the first row immediately after the last row with a given property (ordered by id). Id's may not be consecutive.
Ideally it would look something like this:
...
JOIN (select max(id) id from my_table where CONDITION) m
JOIN (select min(id) from my_table where id > m.id) n
However, I can not use identifier m in the second subselect.
It is possible to use nested queries in nested queries, but is there an easier way?
Thank you.

Answer 1

You could use lead() to get the next id before applying the condition:
select t.*
from my_table t join
(select max(next_id) as max_next_id
from (select t.*, lead(id) over (order by id) as next_id
from my_table t
) t
where <condition>
) tt
on t.id = tt.max_next_id;
You could also do:
select t.*
from my_table t
where t.id > (select max(t2.id) from my_table t2 where <condition>)
order by t2.id asc
fetch first 1 row only;

Answer 2

I am not sure how this is getting woven into the rest of your query, so I have used a CTE
WITH max_next AS (
SELECT r.id as max_id
,r.next_id
FROM (
SELECT m.id
,m.next_id
,ROW_NUMBER() OVER (ORDER BY m.id DESC) AS rn
FROM (
SELECT n.* -- to provide data to satisfy CONDITIONS
,LEAD(n.id) OVER(ORDER BY n.id) as next_id
FROM my_table AS n
) AS m
WHERE CONDITIONS
) AS r
WHERE r.rn = 1
)
I would also shrink the n.* to the columns needed by CONDITIONS to a, not be implicit as the * slows the compile time down (or historically has) as all meta data needs to be read to understand what columns is in the ANY, and the while the compile can also prune not used columns, it's faster if you just ask for what you want (in best case just a compile time savings, worse case, it read all the data when you only need x number of columns read)
And borrowing from Gordon solution, the ROW_NUMBER part could be simpler
WITH max_next AS (
SELECT m.id
,m.next_id
--, plus what ever other things you want from m
FROM (
SELECT n.* -- to satisfy CONDITIONS needs
,LEAD(n.id) OVER(ORDER BY n.id) as next_id
FROM my_table AS n
) AS m
WHERE CONDITIONS
ORDER BY m.id DESC LIMIT 1
)
So for an example for #PIG,
WITH my_table AS (
SELECT column1 AS id
,column2 AS con1
,column3 AS other
FROM VALUES (1,'a',123),(2,'b',234),(3,'a',345),(5,'b',456),(7,'a',567),(10,'c',678)
)
SELECT m.id
,m.next_id
,m.other
FROM (
SELECT n.* -- to satisfy CONDITIONS needs
,LEAD(n.id) OVER(ORDER BY n.id) as next_id
FROM my_table AS n
) AS m
WHERE m.con1 = 'b'
ORDER BY m.id DESC LIMIT 1;
gives 5, 7, 456 which is the last 'b' and the new row, and an extra value on my_table for entertainment purposes (and run on Snowflake to, which means I fixed the prior SQL also.)

Answer 3

This should work, it's pretty straightforward (easy), and it's good that you know records may not be stored in a ordered/consecutive fashion.
SELECT *
FROM my_table
WHERE id = (
SELECT min(id)
FROM my_table
WHERE id > (
SELECT max(id)
FROM my_table
WHERE CONDITION));