How to find the longest sequence of letter in a string - sql-server-2008

Question

I want to find out the longest sequence of letter in a string
e.g. in the word Honorificabcdwert , the output will be abcd.
How can I do it?
My idea is to get the Ascii and then count the sequence until it breaks at some point. But I was able to proceed with only
DECLARE #t TABLE(ID INT IDENTITY,String VARCHAR(100))
INSERT INTO #t SELECT 'Honorificabcdwert'
;with Get_Individual_Chars_Cte AS
(
SELECT
ID
,Row_ID =ROW_NUMBER() Over(PARTITION by ID Order by ID)
,SUBSTRING(String,Number,1) AS [Char]
,ASCII(SUBSTRING(String,Number,1)) AS [Ascii Value]
FROM #t
INNER JOIN master.dbo.spt_values ON
Number BETWEEN 1 AND LEN(String)
AND type='P'
)
Select * from Get_Individual_Chars_Cte
After this I don't know what to do. Help needed for this or any other way of doing so.

Answer 1

Will this help
DECLARE #t TABLE(ID INT IDENTITY,String VARCHAR(100))
INSERT INTO #t
SELECT 'Honorificabcdwert' UNION ALL
SELECT 'AbCdEfxy' UNION ALL
SELECT 'abc1234defg' UNION ALL
SELECT 'XYZABCPPCKLMIDBABC' UNION ALL
SELECT 'MNOP$%^&~()MNOPQRS;:'
SELECT ID, OriginalString,Sequence
FROM (SELECT ID, REPLACE(string,'%','') AS Sequence,OriginalString,
ROW_NUMBER() OVER(PARTITION BY ID ORDER BY LEN(string) DESC, string) AS rn
FROM (SELECT OriginalString = b.String, CASE WHEN b.String LIKE a.strings THEN a.strings ELSE NULL END AS string,
b.ID, ROW_NUMBER() OVER(PARTITION BY ID ORDER BY LEN(strings) DESC, strings) AS rn
FROM (SELECT COALESCE('%' + b.strings+a.strings + '%','%' + a.strings + '%') AS strings
FROM (SELECT SUBSTRING('ABCDEFGHIJKLMNOPQRSTUVWXYZ',t1.N,t2.N-t1.N+1) AS strings, t1.N
FROM (VALUES(1),(2),(3),(4),(5),(6),(7),(8),
(9),(10),(11),(12),(13),(14),(15),
(16),(17),(18),(19),(20),(21),(22),
(23),(24),(25),(26)) t1(N)
CROSS JOIN (VALUES(1),(2),(3),(4),(5),(6),(7),(8),
(9),(10),(11),(12),(13),(14),(15),
(16),(17),(18),(19),(20),(21),(22),
(23),(24),(25),(26)) t2(N)
WHERE t1.N <= t2.N) a
LEFT OUTER JOIN (SELECT REVERSE(SUBSTRING('ZYXWVUTSRQPONMLKJIHGFEDCBA',1,N)) AS strings, 1 AS ID
FROM (VALUES(1),(2),(3),(4),(5),(6),(7),(8),
(9),(10),(11),(12),(13),(14),(15),
(16),(17),(18),(19),(20),(21),(22),
(23),(24),(25),(26)) t1(N)
UNION ALL SELECT '', 1) b ON a.N = b.ID) a
CROSS JOIN #t b) a ) a
WHERE a.rn = 1
ORDER BY a.ID
Result
ID OriginalString Sequence
1 Honorificabcdwert ABCD
2 AbCdEfxy ABCDEF
3 abc1234defg DEFG
4 XYZABCPPCKLMIDBABC XYZABC
5 MNOP$%^&~()MNOPQRS;: MNOPQRS
Based on your inputs provided in the course of discussion with #Martin Smith, the program is being developed. Please test it and let me know if it satisfies your requirement.

Answer 2

For consecutive rows with characters rising in alphabetical order (equating alphabetical order with ASCII order here) ROW_NUMBER() OVER (ORDER BY Row_ID) - [Ascii Value] will be the same.
This is not sufficient on its own however as for the string ABCZE that would put E in the same group as ABC so then you need a second operation to find gaps in that grouping sequence.
Something like the following should do it.
DECLARE #t TABLE(ID INT IDENTITY,String VARCHAR(100))
INSERT INTO #t SELECT 'Honorificabcdwfrt'
;with Get_Individual_Chars_Cte AS
(
SELECT
ID
,Row_ID =ROW_NUMBER() Over(PARTITION by ID Order by ID)
,SUBSTRING(String,number,1) AS [Char]
,ASCII(SUBSTRING(String,number,1)) AS [Ascii Value]
FROM #t
INNER JOIN master.dbo.spt_values ON
number BETWEEN 1 AND LEN(String)
AND type='P'
)
, T1 AS
(
Select *,
ROW_NUMBER() OVER (ORDER BY Row_ID) - [Ascii Value] AS RN
from Get_Individual_Chars_Cte
), T2 AS
(
SELECT *,
ROW_NUMBER() OVER (ORDER BY Row_ID) -
ROW_NUMBER() OVER (PARTITION BY RN ORDER BY Row_ID) AS Grp
FROM T1
)
SELECT TOP 1 WITH TIES *
FROM T2
ORDER BY COUNT(*) OVER (PARTITION BY RN, Grp) DESC