Increment alpha numeric string in stored procedure

What you ask for is NOt an "incrementing" sequence.
You show AA001  coming AFTER  Z0001, and that just doesn't make sense - it wouldn't sort that way...

An incrementing sequence would be

00001-00009-0000A-0000Z
00011-00019-0001A-0001Z
.....
ZZZZ1-ZZZZ9-ZZZZA-ZZZZZ

If you simply must have your original schema, you may need to offer more points to get a solution.

I think you misunderstood, nrbeen, or maybe I am misreading. I see "have a 5 character field" as meaning, 00001 - 99999 is a range. Then the next range is A0001 - A9999.

'00001'
'AA001'
'Z0001'

Alphanumerical sort, 'Z0000' is after 'AA000'. :)

You are correct, the sequence should go like this...
00000
00001

0000A
0000B


0000Z
00010

ZZZZZ

I do need to have a stored procedure to achieve this, I already have a VB.NET code.

Thanks,

I see what you are saying now, nrbreen!

If this helps, but I saw this solution on this site, but it does not take care of the cases where "Z" is between the numbers in the string ( particularly after Z if there is a number ...for example, 000Z0 should goto 000Z1, where as this stored procedure gives an error.)

Any Help is appreciated. Thanks,


CREATE PROCEDURE [dbo].[IncrementCustNum]
      @CustNum VARCHAR(5),
      @NewCustNum VARCHAR(5) OUTPUT
AS
DECLARE @firstChar INT

IF RIGHT(@CustNum, 1) BETWEEN '0' AND '9'
BEGIN
    SET @firstChar = PATINDEX('%[0-9]%', @CustNum)
    IF SUBSTRING(@CustNum, @firstChar, 5) <> LEFT('99999', 6 - @firstChar)
         SET @NewCustNum = LEFT(@CustNum, @firstChar - 1) +
          RIGHT('0000' + CAST(CAST(SUBSTRING(@CustNum, @firstChar, 5) AS INT) + 1 AS VARCHAR(5)), 6 - @firstChar)
    ELSE
         SET @NewCustNum = LEFT(@CustNum, @firstChar - 2) +
          CHAR(ASCII(SUBSTRING(@CustNum, @firstChar - 1, 1)) + 1) +
          RIGHT('0000' + CAST(CAST(SUBSTRING(@CustNum, @firstChar, 5) AS INT) + 1 AS VARCHAR(5)), 6 - @firstChar)
END --IF
ELSE
BEGIN
IF RIGHT(@CustNum, 1) < 'Z'
    SET @NewCustNum = LEFT(@CustNum, LEN(@CustNum) - 1) + CHAR(ASCII(RIGHT(@CustNum, 1)) + 1)
ELSE
      BEGIN
            SET @firstChar = LEN(@CustNum) + 1 - PATINDEX('%[^Z]%', REVERSE(@CustNum))
            IF @firstChar > LEN(@CustNum)  -- @CustNum is all Z's!!
                  SET @NewCustNum = '00000'
            ELSE
                SET @NewCustNum = LEFT(@CustNum, @firstChar - 1) +
                        CHAR(ASCII(SUBSTRING(@CustNum, @firstChar, 1)) + 1) +
                        REPLICATE('0', 5 - @firstChar)
      END --ELSE
END --ELSE

GO

I am sure this can be done some easier way, but here is what comes to my mind.

-Create a table of valid values, '00000' to 'ZZZZZ', in a single CHAR(5) column. e.g., CustNums
-Index this column.
-In your increment customer number code, simply grab the MIN(CustNum) from your CustNums table WHERE CustNum > @CustNum

Hi mwvisa1, This table will need few million values, so it's not a feasible solution.

Why NOT? You only have to create the table once and with the index, it will probably perform better than the stored procedure and with much less code.

This should give the full range of increments -
extra vars used to simplify debugging.

CREATE PROCEDURE [dbo].[Z_Increment]
      @inputcode VARCHAR(5),
      @result VARCHAR(5) OUTPUT
AS

declare @i as int , @c as char(1), @c2 as char(1), @cval as int, @cval2 as int, @carry as int 

select @result=''
select @carry=1
select @i=5
while @i>0
  begin
    select @c=substring(@inputcode,@i,1)
    select @cval=ascii(@c)   
    
    select @cval2=@cval + @carry
    select @carry=0
    
    select @c2=CHAR(@cval2)
    if @cval2=10 select @c2='A'
    if @cval2 =91 
       begin
       select @c2='0'
       select @carry=1
       end

    select @result=@c2  + @result   -- append to front of result 
    select @i=@i-1
  end
-- select @result 

Select allOpen in new window

@dotneteng: I guess it is 60 million. *laughing* Should have done the math. :)

@nrbreen: looks good. If you are using ASCII() value, though, when is the @cval2 going to equal 10? Do you mean 58.

0-9 is ASCII 48-57, A-Z is 65-90; therefore, the 91 makes sense, just not the 10.

nrbreen,

This is almost correct. It doesn't work with the cases where there is 9 at the end...
00999, 00099...I am stilll testing with some more cases.

Thanks!

Replace:
if @cval2=10 select @c2='A'

With:
if @cval2=58 select @c2='A'

And sorry again @dotneteng about the not doing math on number of rows. I have utility tables with million rows like numbers table and performs quite well, but sixty million is another ball game. *smile*