SQL issue on query. need a better/alternate query - sql

Question

Say we have a table T
+------+
| NUMM |
+------+
| 1 |
| 5 |
| 3 |
| 8 |
+------+
I want the nearest bigger number from the column numm to be in column numm1.
The result will look like this
+-------------+
| NUMM | NUMM1|
+-------------+
| 1 | 3 |
| 3 | 5 |
| 5 | 8 |
+-------------+
I wrote a query like this and it works. But i would like to know if there is a better way for sollution.
select numm, numm + min(dif) as numm1
from (select distinct a.numm numm, b.numm numm1, b.numm - a.numm dif
from (select *
from T
where numm != (select max(numm) from T )) a
join T b
on 1 = 1)
where dif > 0
group by numm

Answer 1

If you want to get the direct successor, you can use the lead() windowing function:
select * from (
select
numm,
lead(numm) over (order by numm) as numm1
from t
)
where numm1 is not null
order by numm;

Answer 2

If it's oracle, you can use row_number() function to rank, then inner join with [left_table].rank = [right_table].rank - 1:
SELECT a.numm,
b.numm
FROM
(SELECT numm, row_number() over(order by numm) AS rank FROM pn_test) a
INNER JOIN
(SELECT numm, row_number() over(order by numm) AS rank FROM pn_test) b
ON a.rank = b.rank - 1;

Answer 3

Try the below query if you have repeated records in your table:
WITH CTE_ABC
AS (
SELECT DISTINCT NUMM
FROM [Table]
)
,CTE_XYZ
AS (
SELECT *
FROM (
SELECT NUMM
,lead(NUMM) OVER (
ORDER BY NUMM
) AS numm1
FROM CTE_ABC
) A
WHERE numm1 IS NOT NULL
)
SELECT A.NUMM
,B.numm1
FROM [Table] A
LEFT JOIN CTE_XYZ B
ON A.columnId = B.columnId
WHERE B.numm1 IS NOT NULL

Answer 4

Frank's answer is probably the best when there are no duplicate numbers, but if you can end up with duplicates here's one possible solution:
with t1 as (
select numm
, dense_rank() over (order by numm) rnk
from t
)
select t1.numm
, t2.numm numm1
from t1
join (select distinct numm, rnk-1 rnk from t1) t2
on t1.rnk = t2.rnk;
In this solution the DENSE_RANK analytic function is first used in T1 to give every distinct NUMM a sequential number (RNK). In the second stage T1 is join on RNK to the distinct set of numm and rnk-1 values from t1.
Looking for better performance, this might do the job:
with t1 as (
select numm
, dense_rank() over (order by numm) rnk
, row_number() over (partition by numm order by rownum) ord
from t
)
select t1.numm
, t2.numm numm1
from t1
join t1 t2
on t1.rnk = t2.rnk-1
and t2.ord = 1;
Here I added a way to grab one record for each numm to the subfactored query and eliminated the distinct operation in T2. With a limited data set of just 5 records and no indexes it has a cost of 9 vs a cost of 10 for the prior query.