Lag and Lead to next month - oracle

Question

TABLE: HIST
CUSTOMER MONTH PLAN
1 1 A
1 2 B
1 2 C
1 3 D
If I query:
select h.*, lead(plan) over (partition by customer order by month) np from HIST h
I get:
CUSTOMER MONTH PLAN np
1 1 A B
1 2 B C
1 2 C D
1 3 D (null)
But I wanted
CUSTOMER MONTH PLAN np
1 1 A B
1 2 B D
1 2 C D
1 3 D (null)
Reason being, next month to 2 is 3, with D. I'm guessing partition by customer order by month doesn't work the way I thought.
Is there a way to achieve this in Oracle 12c?

Answer 1

One way to do it is to use RANGE partitioning with the MIN analytic function. Like this:
select h.*,
min(plan) over
(partition by customer
order by month
range between 1 following and 1 following) np
from HIST h;
+----------+-------+------+----+
| CUSTOMER | MONTH | PLAN | NP |
+----------+-------+------+----+
| 1 | 1 | A | B |
| 1 | 2 | B | D |
| 1 | 2 | C | D |
| 1 | 3 | D | |
+----------+-------+------+----+
When you use RANGE partitioning, you are telling Oracle to make the windows based on the values of the column you are ordering by rather than making the windows based on the rows.
So, e.g.,
ROWS BETWEEN 1 following and 1 following
... will make a window containing the next row.
RANGE BETWEEN 1 following and 1 following
... will make a window containing all the rows having the next value for month.
UPDATE
If it is possible that some values for MONTH might be skipped for a given customer, you can use this variant:
select h.*,
first_value(plan) over
(partition by customer
order by month
range between 1 following and unbounded following) np
from h
+----------+-------+------+----+
| CUSTOMER | MONTH | PLAN | NP |
+----------+-------+------+----+
| 1 | 1 | A | B |
| 1 | 3 | B | D |
| 1 | 3 | C | D |
| 1 | 4 | D | |
+----------+-------+------+----+

Answer 2

You can use LAG/LEAD twice. The first time to check for duplicate months and to set the value to NULL in those months and the second time use IGNORE NULLS to get the next monthly value.
It has the additional benefit that if months are skipped then it will still find the next value.
SQL Fiddle
Oracle 11g R2 Schema Setup:
CREATE TABLE HIST ( CUSTOMER, MONTH, PLAN ) AS
SELECT 1, 1, 'A' FROM DUAL UNION ALL
SELECT 1, 2, 'B' FROM DUAL UNION ALL
SELECT 1, 2, 'C' FROM DUAL UNION ALL
SELECT 1, 3, 'D' FROM DUAL UNION ALL
SELECT 2, 1, 'E' FROM DUAL UNION ALL
SELECT 2, 1, 'F' FROM DUAL UNION ALL
SELECT 2, 3, 'G' FROM DUAL UNION ALL
SELECT 2, 5, 'H' FROM DUAL;
Query 1:
SELECT CUSTOMER,
MONTH,
PLAN,
LEAD( np ) IGNORE NULLS OVER ( PARTITION BY CUSTOMER ORDER BY MONTH, PLAN, ROWNUM ) AS np
FROM (
SELECT h.*,
CASE MONTH
WHEN LAG( MONTH ) OVER ( PARTITION BY CUSTOMER ORDER BY MONTH, PLAN, ROWNUM )
THEN NULL
ELSE PLAN
END AS np
FROM hist h
)
Results:
| CUSTOMER | MONTH | PLAN | NP |
|----------|-------|------|--------|
| 1 | 1 | A | B |
| 1 | 2 | B | D |
| 1 | 2 | C | D |
| 1 | 3 | D | (null) |
| 2 | 1 | E | G |
| 2 | 1 | F | G |
| 2 | 3 | G | H |
| 2 | 5 | H | (null) |

Answer 3

Just so that it is listed here as an option for Oracle 12c (onward), you can use an apply operator for this style of problem
select
h.customer, h.month, h.plan, oa.np
from hist h
outer apply (
select
h2.plan as np
from hist h2
where h.customer = h.customer
and h2.month > h.month
order by month
fetch first 1 rows only
) oa
order by
h.customer, h.month, h.plan
I don't know of any Oracle 12c public fiddles so, an example in SQL Server can be found here: http://sqlfiddle.com/#!18/cd95e/1
| customer | month | plan | np |
|----------|-------|------|--------|
| 1 | 1 | A | C |
| 1 | 2 | B | D |
| 1 | 2 | C | D |
| 1 | 3 | D | (null) |