sql extract rightmost number in string and increment - sql

Question

i have transaction codes like
"A0004", "1B2005","20CCCCCCC21"
I need to extract the rightmost number and increment the transaction code by one
"AA0004"----->"AA0005"
"1B2005"------->"1B2006"
"20CCCCCCCC21"------>"20CCCCCCCC22"
in SQL Server 2012.
unknown length of string
right(n?) always number
dealing with unsignificant number of string and number length is out of my league.
some logic is always missing.
LEFT(#a,2)+RIGHT('000'+CONVERT(NVARCHAR,CONVERT(INT,SUBSTRING( SUBSTRING(#a,2,4),2,3))+1)),3

Answer 1

First, I want to be clear about this: I totally agree with the comments to the question from a_horse_with_no_name and Jeroen Mostert.
You should be storing one data point per column, period.
Having said that, I do realize that a lot of times the database structure can't be changed - so here's one possible way to get that calculation for you.
First, create and populate sample table (Please save us this step in your future questions):
DECLARE #T AS TABLE
(
col varchar(100)
);
INSERT INTO #T (col) VALUES
('A0004'),
('1B2005'),
('1B2000'),
('1B00'),
('20CCCCCCC21');
(I've added a couple of strings as edge cases you didn't mention in the question)
Then, using a couple of cross apply to minimize code repetition, I came up with that:
SELECT col,
LEFT(col, LEN(col) - LastCharIndex + 1) +
REPLICATE('0', LEN(NumberString) - LEN(CAST(NumberString as int))) +
CAST((CAST(NumberString as int) + 1) as varchar(100)) As Result
FROM #T
CROSS APPLY
(
SELECT PATINDEX('%[^0-9]%', Reverse(col)) As LastCharIndex
) As Idx
CROSS APPLY
(
SELECT RIGHT(col, LastCharIndex - 1) As NumberString
) As NS
Results:
col Result
A0004 A0005
1B2005 1B2006
1B2000 1B2001
1B00 1B01
20CCCCCCC21 20CCCCCCC22
The LastCharIndex represents the index of the last non-digit char in the string.
The NumberString represents the number to increment, as a string (to preserve the leading zeroes if they exists).
From there, it's simply taking the left part of the string (that is, up until the number), and concatenate it to a newly calculated number string, using Replicate to pad the result of addition with the exact number of leading zeroes the original number string had.

Answer 2

Try This
DECLARE #test nvarchar(1000) ='"A0004", "1B2005","20CCCCCCC21"'
DECLARE #Temp AS TABLE (ID INT IDENTITY,Data nvarchar(1000))
INSERT INTO #Temp
SELECT #test
;WITH CTE
AS
(
SELECT Id,LTRIM(RTRIM((REPLACE(Split.a.value('.' ,' nvarchar(max)'),'"','')))) AS Data
,RIGHT(LTRIM(RTRIM((REPLACE(Split.a.value('.' ,' nvarchar(max)'),'"','')))),1)+1 AS ReqData
FROM
(
SELECT ID,
CAST ('<S>'+REPLACE(Data,',','</S><S>')+'</S>' AS XML) AS Data
FROM #Temp
) AS A
CROSS APPLY Data.nodes ('S') AS Split(a)
)
SELECT CONCAT('"'+Data+'"','-------->','"'+CONCAT(LEFT(Data,LEN(Data)-1),CAST(ReqData AS VARCHAR))+'"') AS ExpectedResult
FROM CTE
Result
ExpectedResult
-----------------
"A0004"-------->"A0005"
"1B2005"-------->"1B2006"
"20CCCCCCC21"-------->"20CCCCCCC22"

Answer 3

STUFF(#X
,LEN(#X)-CASE PATINDEX('%[A-Z]%',REVERSE(#X)) WHEN 0 THEN LEN(#X) ELSE PATINDEX('%[A-Z]%',REVERSE(#X))-1 END+1
,LEN(((RIGHT(#X,CASE PATINDEX('%[A-Z]%',REVERSE(#X)) WHEN 0 THEN LEN(#X) ELSE PATINDEX('%[A-Z]%',REVERSE(#X))-1 END)/#N)+1)#N)
,((RIGHT(#X,CASE PATINDEX('%[A-Z]%',REVERSE(#X)) WHEN 0 THEN LEN(#X) ELSE PATINDEX('%[A-Z]%',REVERSE(#X))-1 END)/#N)+1)#N)
works on number only strings
99 becomes 100
mod(#N) increments