Many thanks Chip - no problem - the number of increments is unlikely to
exceed half a dozen.

D10-001 etc are door numbers in a schedule - the alpha suffix is only used
when a door is lotted in later.
I just wanted to be able to slot in a few more is necessary. If more than a
few, total renumbering would be justified.

I am most grateful

Francis


"Chip Pearson" wrote in message
...
Francis,

The following formula and VBA function will incrment A - Z - a - z.
What should happen if the last character is a lower case 'z'?


=LEFT(A1,LEN(A1)-1)&CHAR(CODE(RIGHT(A1,1))+1+(6*(RIGHT(A1,1)="Z")) )


Function IncrAlpha(InChars As String) As String
IncrAlpha = Left(InChars, Len(InChars) - 1) & _
Chr(Asc(Right(InChars, 1)) + 1 - (6 * (Right(InChars, 1) = "Z")))
End Function


--
Cordially,
Chip Pearson
Microsoft MVP - Excel, 10 Years
Pearson Software Consulting
www.cpearson.com
(email on the web site)


"Francis Hookham" wrote in message
...
Earlier postings have shown me how to increment a string where the last
characters are numerals so

D01-004

can be incremented to

D01-005

(see code below)



I also need to be able to increment the last character when it is a
letter rather than a number, for instance,

D01-006c

incremented to

D01-006d



Any suggestions?



Francis Hookham



Code used to increment D01-004 to D01-005 is

sZeroes = "0000000000"

sSuffix = Right(sDNum, Len(sDNum) - InStr(sDNum, "-"))

sDNum = Left(sDNum, InStr(sDNum, "-")) & _

Format(sSuffix + 1, Left(sZeroes, Len(sSuffix)))