Select element show up in the next 2 month. SQL SERVER - sql-server

Question

This is a result of a SELECT query of mine.
Code Name MONTH
------- ------------ -----------
Cust3 CustName3 1
Cust2 CustName2 2
Cust5 CustName5 3
Cust2 CustName2 5
Cust2 CustName2 6
Cust3 CustName3 7
Cust6 CustName6 8
Cust1 CustName1 10
Cust1 CustName1 11
Cust3 CustName3 12
Now I want to select rows that have the same Code value for two consecutive months, for example Cust2 is contiguous in months 5 and 6 and Cust1 is contiguous in months 10 and 11:
Code Name MONTH
------- ------------ -----------
Cust2 CustName2 5
Cust2 CustName2 6
Cust1 CustName1 10
Cust1 CustName1 11

Answer 1

Here is another approach using ROW_NUMBER and COUNT() OVER:
SQL Fiddle
WITH Cte AS(
SELECT *,
RN = [Month] - ROW_NUMBER() OVER(PARTITION BY Code ORDER BY [Month])
FROM TestData
),
Cte2 AS(
SELECT *,
CC = COUNT(*) OVER(PARTITION BY Code, RN)
FROM Cte
)
SELECT
Code, Name, [Month]
FROM Cte2
WHERE CC >= 2

Answer 2

I assume the underlying table is denormalized and has this schema:
Code varchar(n),
Name nvarchar(n),
Month int
...and that there is only 1 row for each month.
We can use the LAG function in T-SQL to compare neighbour rows. We don't need the optional OVER partition-by clause as it's a single data set group:
SELECT
[Code],
[Month]
FROM
(
SELECT
[Code],
LAG( [Code], 1 ) AS LastCode
[Month],
LAG( [Month], 1 ) AS LastMonth
FROM
[sourceTable]
ORDER BY
[Month] ASC
) AS Lagged
WHERE
Lagged.[Code] = Lagged.LastCode
AND
Lagged.[Month] = Lagged.LastMonth

Answer 3

The above two answers are pretty neat. Here's another alternative way of solving this (example), perhaps not as elegantly:
select * from
(
-- join to itself, but for month - join to previous month
select a.code, a.month
from test a
inner join test b on a.code=b.code and a.month = b.month-1
union all
-- join to itself, but for month - join to next month
select a.code, a.month
from test a
inner join test b on a.code=b.code and a.month = b.month+1
) a
order by a.code, a.month